In this post we have provided Solutions for NCERT Class 12 Maths Integration Miscellaneous Exercise. This exercise covers all the topics covered through chapter 7, this makes it essential for mastering integration techniques for board exams.
Here, we’ll cover step-by-step solutions for Questions of the Miscellaneous Exercise from Chapter 7: Integrals. These will be helpful for preparing for both your BOARDs and Competetive exams.
Question 1: Integrate \int \frac{1}{x-x^{3}} \ dx
Solution:
I = \int \frac{1}{x-x^{3}} \ dx
I = \int \frac{1}{x(1-x^{2})} \ dx
\frac{1}{x(1-x^{2})} = \frac{A}{x}+\frac{Bx+C}{1-x^{2}}
1 = A(1-x^{2})+(Bx+C)x
1 = A - A x^{2} + Bx^{2}+Cx
\text{Equating coefficients: } A=1,\ -A+B=0 \implies B=1,\ C=0
\frac{1}{x(1-x^{2})} = \frac{1}{x}+\frac{x}{1-x^{2}}
I = \int \frac{1}{x} \ dx + \int \frac{x}{1-x^{2}} \ dx
I= \ln|x| - \tfrac{1}{2}\ln|1-x^{2}|+C
I= \frac{1}{2}\log|x^2| - \frac{1}{2}\log|1-x^{2}|+C
I= \frac{1}{2}\log \left| \frac{x^2}{1-x^{2}} \right| + C
Final Answer: \boxed{\int \frac{1}{x-x^{3}} \ dx = \frac{1}{2}\log \left| \frac{x^2}{1-x^{2}} \right| + C}
Question 2: Integrate \int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} \ dx
Solution:
I = \int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} \ dx
\text{Multiply numerator and denominator by } \sqrt{x+a}-\sqrt{x+b}
I = \int \frac{\sqrt{x+a}-\sqrt{x+b}}{(\sqrt{x+a}+\sqrt{x+b})(\sqrt{x+a}-\sqrt{x+b})} \ dx
I = \int \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} \ dx
= \int \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b} \ dx
= \frac{1}{a-b}\int \sqrt{x+a} \ dx - \frac{1}{a-b}\int \sqrt{x+b} \ dx
\int \sqrt{x+a} \ dx = \frac{2}{3}(x+a)^{3/2}, \quad \int \sqrt{x+b} \ dx = \frac{2}{3}(x+b)^{3/2}
I = \frac{2}{3(a-b)}\big((x+a)^{3/2}-(x+b)^{3/2}\big)+C
Final Answer: \boxed{\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} \ dx = \frac{2}{3(a-b)}\Big((x+a)^{3/2}-(x+b)^{3/2}\Big)+C}
Question 3: Integrate \int \frac{1}{x\sqrt{ax-x^{2}}} \ dx
Solution:
Let, I=\int \frac{1}{x\sqrt{ax-x^{2}}} \ dx
\text{Put }x=\frac{a}{t}\Rightarrow dx=-\frac{a}{t^{2}} \ dt
\sqrt{ax-x^{2}}=\sqrt{a\cdot \frac{a}{t}-\frac{a^{2}}{t^{2}}}=a\sqrt{\frac{t-1}{t^{2}}}=\frac{a\sqrt{t-1}}{t}
\text{Hence} I = \int \frac{1}{(a/t)\cdot (a\sqrt{t-1}/t)}\cdot\left(-\frac{a}{t^{2}} dt\right)
I=- \int \frac{1}{a}\cdot\frac{1}{\sqrt{t-1}} \ dt
I=-\frac{1}{a}\int (t-1)^{-1/2} \ dt
I =-\frac{1}{a}\cdot 2\sqrt{t-1}+C
\text{Back-substitute }t=\frac{a}{x}
\sqrt{t-1}=\sqrt{\frac{a}{x}-1}=\sqrt {\frac{a-x}{x}}
I = -\frac{2}{a}\sqrt{\frac{a-x}{x}}+C
Final Answer: \boxed{ \int \frac{1}{x\sqrt{ax-x^{2}}},dx=-\frac{2}{a}\sqrt{\frac{a-x}{x}}+C}
Question 4: Integrate \int \frac{1}{x^{2}\bigl(x^{4}+1\bigr)^{3/4}} \ dx
Solution:
Let, I=\int \frac{1}{x^{2}(x^{4}+1)^{3/4}} \ dx
I=\int \frac{1}{x^{2}x^{3}(1+\frac{1}{x^{4}})^{3/4}} \ dx
I=\int \frac{1}{x^{5}(1+\frac{1}{x^{4}})^{3/4}} \ dx
Let, t = 1 + \frac{1}{x^4} \implies dt = -\frac{4dx}{x^5} \implies -\frac{dt}{4} = \frac{dx}{x^5}
I = \frac{-1}{4}\int \frac{dt}{t^{\tfrac{3}{4}}}
I = \frac{-1}{4}\int t^{-\tfrac{3}{4}}dt
I = \frac{-1}{4} \frac{t^{\tfrac{1}{4}}}{\frac{1}{4}} + C
Substituting back t
I = - (1 + \frac{1}{x^4})^{\tfrac{1}{4}} + C
Final Answer: \boxed{ \int \frac{1}{x^{2}(x^{4}+1)^{3/4}} \ dx = - \left(1 + \frac{1}{x^4} \right)^{\frac{1}{4}} + C }
Question 5: Integrate \int \frac{1}{x^{1/2}+x^{1/3}} \ dx
Solution:
Let, I=\int \frac{1}{x^{1/2}+x^{1/3}} \ dx
\text{Put }x=t^{6}\Rightarrow dx=6t^{5} dt, \ x^{1/2}=t^{3},\ x^{1/3}=t^{2}
I = \int \frac{1}{t^{3}+t^{2}}\cdot 6t^{5} \ dt
I=6\frac{t^{5}}{t^{2}(t+1)} \ dt
I =6\frac{t^{3}}{t+1} \ dt
(\text{using polynomial division}) \quad \frac{t^{3}}{t+1}=t^{2}-t+1-\frac{1}{t+1}
I=6\int\Big(t^{2}-t+1-\frac{1}{t+1}\Big) \ dt
I=6\Big(\frac{t^{3}}{3}-\frac{t^{2}}{2}+t-\ln|t+1|\Big)+C
I=2t^{3}-3t^{2}+6t-6\ln|t+1|+C
\text{Back-substitute }t=x^{1/6}:
I = 2x^{1/2}-3x^{1/3}+6x^{1/6}-6\ln\bigl(1+x^{1/6}\bigr)+C
Final Answer: \boxed{ \int \frac{1}{x^{1/2}+x^{1/3}} \ dx=2x^{1/2}-3x^{1/3}+6x^{1/6}-6\ln\bigl(1+x^{1/6}\bigr)+C}
Question 6: Integrate \int \frac{5x}{(x+1)(x^{2}+9)} \ dx
Solution:
Let, I=\int \frac{5x}{(x+1)(x^{2}+9)} \ dx
\text{Partial fractions: }\frac{5x}{(x+1)(x^{2}+9)}=\frac{A}{x+1}+\frac{Bx+C}{x^{2}+9}
5x=A(x^{2}+9)+(Bx+C)(x+1)
5x=A x^{2}+9A + Bx^{2}+ (B+C)x + C
\text{Collect: }x^{2}:(A+B),\ x:(B+C),\ \text{const }:(9A+C)
\text{Compare with }5x:\ A+B=0,\ B+C=5,\ 9A+C=0
\text{From }A+B=0\Rightarrow B=-A.\ \text{From }9A+C=0\Rightarrow C=-9A
B+C=-A-9A=-10A=5\Rightarrow A=-\tfrac{1}{2}\Rightarrow B=\tfrac{1}{2},\ C=\tfrac{9}{2}
\therefore \frac{5x}{(x+1)(x^{2}+9)}=-\frac{1/2}{x+1}+\frac{\frac{1}{2}x+\frac{9}{2}}{x^{2}+9}
I= -\frac{1}{2}\int\frac{1}{x+1} \ dx+\frac{1}{2}\int\frac{x}{x^{2}+9} \ dx+\frac{9}{2}\int\frac{1}{x^{2}+9} \ dx
\int\frac{x}{x^{2}+9} \ dx = \frac{1}{2}\ln(x^{2}+9) , \quad \int\frac{1}{x^{2}+9} \ dx=\frac{1}{3}\tan^{-1} \frac{x}{3}
I=-\frac{1}{2}\ln|x+1|+\frac{1}{2} \cdot \frac{1}{2}\ln(x^{2}+9)+\frac{9}{2}\cdot\frac{1}{3}\tan^{-1}\frac{x}{3}+C
I=-\frac{1}{2}\ln|x+1|+\frac{1}{4}\ln(x^{2}+9)+\frac{3}{2}\tan^{-1}\frac{x}{3}+C
Final Answer: \boxed{\int \frac{5x}{(x+1)(x^{2}+9)} \ dx=-\frac{1}{2}\ln|x+1|+\frac{1}{4}\ln(x^{2}+9)+\frac{3}{2}\tan^{-1}\frac{x}{3}+C}
Question 7: Integrate \int \frac{\sin x}{\sin(x-a)} \ dx
Solution:
Let, I=\int \frac{\sin x}{\sin(x-a)},dx
\text{Put }u=x-a\Rightarrow x=u+a \implies dx=du
\sin x=\sin(u+a)=\sin u\cos a+\cos u\sin a
\frac{\sin x}{\sin(x-a)}=\frac{\sin u\cos a+\cos u\sin a}{\sin u}=\cos a+\sin a\cot u
I=\int \cos a \ du+\sin a\int \cot u \ du
I=\cos a\cdot u+\sin a\ln|\sin u|+C
\text{Back-substitute }u=x-a:
I=\cos a\cdot (x-a)+\sin a\ln\bigl|\sin(x-a)\bigr|+C
I=x\cos a - a\cos a+\sin a\ln\bigl|\sin(x-a)\bigr|+C
I=x\cos a +\sin a\ln\bigl|\sin(x-a)\bigr|+C
Final Answer: \boxed{\int \frac{\sin x}{\sin(x-a)} \ dx=x\cos a +\sin a\ln\bigl|\sin(x-a)\bigr|+C}
Question 8: Integrate \int \frac{e^{5\log x}-e^{4\log x}}{e^{3\log x}-e^{2\log x}} \ dx
Solution:
Let, I = \int \frac{e^{5\log x}-e^{4\log x}}{e^{3\log x}-e^{2\log x}} \ dx
We Know, e^{k\log x}=x^{k}
\therefore \frac{e^{5\log x}-e^{4\log x}}{e^{3\log x}-e^{2\log x}} = \frac{x^{5}-x^{4}}{x^{3}-x^{2}}=\frac{x^{4}(x-1)}{x^{2}(x-1)}=x^{2}\ (x\neq 0,1)
I =\int x^{2} dx
I =\frac{x^{3}}{3}+C
Final Answer: \boxed{\int \frac{e^{5\log x}-e^{4\log x}}{e^{3\log x}-e^{2\log x}} \ dx=\frac{x^{3}}{3}+C}
Question 9: Integrate \int \frac{\cos x}{\sqrt{4-\sin^{2}x}} \ dx
Solution:
Let, I=\int \frac{\cos x}{\sqrt{4-\sin^{2}x}} \ dx
\text{Put }t=\sin x\Rightarrow dt=\cos x dx
Substituting in I
I=\int \frac{1}{\sqrt{4-t^{2}}} \ dt
I=\sin^{-1}\frac{t}{2}+C
Final Answer: \boxed{ \int \frac{\cos x}{\sqrt{4-\sin^{2}x}} \ dx=\sin^{-1} \biggl( \frac{\sin x}{2} \biggr)+C}
Question 10: Integrate \int \frac{\sin^{8}x-\cos^{8}x}{1-2\sin^{2}x\cos^{2}x} \ dx
Solution:
Let, \int \frac{\sin^{8}x-\cos^{8}x}{1-2\sin^{2}x\cos^{2}x} \ dx
\sin^{8}x-\cos^{8}x= (\sin^{4} x - \cos^{4} x)( \sin^{4} x+ \cos^{4} x)
\sin^{8}x-\cos^{8}x = (\sin^{2} x - \cos^{2} x)(\sin^{2} x+\cos^{2} x)(\sin^{4} x+\cos^{4} x)
\sin^{8}x-\cos^{8}x = (\sin^{2} x - \cos^{2} x)\cdot 1\cdot (\sin^{4} x+ \cos^{4} x)
\sin^{8}x-\cos^{8}x = (\sin x^{2}-\cos^{2} x)(\sin^{4} x+\cos^{4} x)
2\sin^{2}x\cos^{2}x = (\sin^{2} x + \cos^{2} x)^2 - \sin^{4} x - \cos^{4} x \quad(\text{since }2ab = (a+b)^2 - a^2 -b^2)
2\sin^{2}x\cos^{2}x = 1 - (\sin^{4} x + \cos^{4} x)
1-2\sin^{2}x\cos^{2}x= 1 - (1 - (\sin^{4} x + \cos^{4} x))
1-2\sin^{2}x\cos^{2}x= (\sin^{4} x + \cos^{4} x)
Therefore, \frac{\sin^{8}x-\cos^{8}x}{1-2\sin^{2}x\cos^{2}x} = \frac{(\sin x^{2}-\cos^{2} x)(\sin^{4} x+\cos^{4} x)}{(\sin^{4} x + \cos^{4} x)}
OR \frac{\sin^{8}x-\cos^{8}x}{1-2\sin^{2}x\cos^{2}x} = \sin^{2}x-\cos^{2}x
I=\int (\sin^{2}x-\cos^{2}x) \ dx
I=\int (-\cos 2x) \ dx
I=-\frac{1}{2}\sin 2x +C
Final Answer: \boxed{ \int \frac{\sin^{8}x-\cos^{8}x}{1-2\sin^{2}x\cos^{2}x} \ dx=-\frac{1}{2}\sin 2x +C}
Question 11: Integrate \int \frac{1}{\cos(x+a)\cos(x+b)} \ dx
Solution:
I=\int \frac{1}{\cos(x+a)\cos(x+b)} \ dx
Dividing and multiplying by \sin(a-b)
I= \frac{1}{\sin(a-b)} \int \frac{\sin(a-b)}{\cos(x+a)\cos(x+b)} \ dx
I= \frac{1}{\sin(a-b)} \int \frac{\sin(a-b+x-x)}{\cos(x+a)\cos(x+b)} \ dx
I= \frac{1}{\sin(a-b)} \int \frac{\sin\big( (x+a) - (x+b) \big) }{\cos(x+a)\cos(x+b)} \ dx
I= \frac{1}{\sin(a-b)} \int \frac{\sin(x+a) \cos(x+b) - \cos(x+a) \sin(x+b) }{\cos(x+a)\cos(x+b)} \ dx
I= \frac{1}{\sin(a-b)} \int \frac{\sin(x+a) \cos(x+b) - \cos(x+a) \sin(x+b) }{\cos(x+a)\cos(x+b)} \ dx
I= \frac{1}{\sin(a-b)} \int \frac{\sin(x+a) \cos(x+b)}{\cos(x+a)\cos(x+b)} - \frac{\cos(x+a) \sin(x+b) \big)}{\cos(x+a)\cos(x+b)} \ dx
I= \frac{1}{\sin(a-b)} \int \frac{\sin(x+a)}{\cos(x+a)} - \frac{\sin(x+b) \big)}{\cos(x+b)} \ dx
I= \frac{1}{\sin(a-b)} \int \tan(x+a) - \tan(x+b) \ dx
I= \frac{1}{\sin(a-b)} \big[ -\log |\cos(x+a)| + \log |\cos(x+b)| \big] + C
I= \frac{1}{\sin(a-b)} \big[\log \left| \frac{\cos(x+b)}{\cos(x+a)} \right| \big] + C
Final Answer: \boxed{ \int \frac{1}{\cos(x+a)\cos(x+b)} \ dx= \frac{1}{\sin(a-b)} \big[\log \left| \frac{\cos(x+b)}{\cos(x+a)} \right| \big] + C }
Question 12: Integrate \int \frac{x^{3}}{\sqrt{1-x^{8}}} \ dx
Solution:
Let, I = \int \frac{x^{3}}{\sqrt{1-x^{8}}} \ dx
I=\int \frac{x^{3}}{\sqrt{1-x^{8}}} \ dx
\text{Put }t=x^{4}\Rightarrow dt=4x^{3}dx\Rightarrow x^{3}dx=\frac{dt}{4}
1-x^{8}=1-t^{2}
I =\tfrac{1}{4}\int \frac{1}{\sqrt{1-t^{2}}} \ dt
I =\frac{1}{4}\sin^{-1} t + C
Final Answer: \boxed{\int \frac{x^{3}}{\sqrt{1-x^{8}}} \ dx = \frac{1}{4}\sin^{-1} \bigl( x^{4} \bigr)+C}
Question 13: Integrate \int \frac{e^{x}}{(1+e^{x})(2+e^{x})} \ dx
Let, I=\int \frac{e^{x}}{(1+e^{x})(2+e^{x})} \ dx
\text{Put }t=e^{x}\Rightarrow dt=e^{x}dx
I=\int \frac{1}{(1+t)(2+t)},dt
Using Partial Fractions
\frac{1}{(t+1)(t+2)}=\frac{A}{t+1}+\frac{B}{t+2}
1=A(t+2)+B(t+1) \Rightarrow A+B=0,\ 2A+B=1\Rightarrow A=1,\ B=-1
\frac{1}{(t+1)(t+2)}=\frac{1}{t+1}-\frac{1}{t+2}
I=\int\left(\frac{1}{t+1}-\frac{1}{t+2}\right) \ dt
I=\log|t+1|-\log|t+2|+C
I=\log|\frac{t+1}{t+2}|+C
I=\log|\frac{e^{x}+1}{e^{x}+2}|+C
Final Answer: \boxed{\int\frac{e^{x}}{(1+e^{x})(2+e^{x})},dx=\log \left| \frac{e^{x}+1}{e^{x}+2} \right|+C}
Question 14: Integrate \int \frac{1}{(x^{2}+1)(x^{2}+4)} \ dx
Solution:
Let I=\int \frac{1}{(x^{2}+1)(x^{2}+4)} \ dx
Using Partial Fraction
\frac{1}{(x^{2}+1)(x^{2}+4)}=\frac{A}{x^{2}+1}+\frac{B}{x^{2}+4}
1=A(x^{2}+4)+B(x^{2}+1)
1=(A+B)x^{2}+4A+B
A+B=0,\quad 4A+B=1\Rightarrow A=\tfrac{1}{3},\ B=-\tfrac{1}{3}
I=\frac{1}{3}\int\frac{1}{x^{2}+1} \ dx - \frac{1}{3}\int\frac{1}{x^{2}+4} \ dx
I=\frac{1}{3}\tan^{-1} x-\tfrac{1}{3}\cdot\frac{1}{2}\tan^{-1} \frac{x}{2}+C
Final Answer: \boxed{ \int\frac{1}{(x^{2}+1)(x^{2}+4)},dx=\frac{1}{3}\tan^{-1} x-\frac{1}{6}\tan^{-1} \frac{x}{2}+C}
Question 15: Integrate \int \cos^{3}x\ e^{\log\sin x} \ dx
Solution:
Let, I = \int \cos^{3}x\ e^{\log\sin x} \ dx
We know
e^{\log\sin x}=\sin x\ \quad (\text{assume }\sin x>0\text{ on domain})
I=\int \cos^{3}x\cdot\sin x \ dx
\text{Let }u=\cos x\Rightarrow du=-\sin x dx
I=\int -u^{3} \ du
I= -\frac{u^{4}}{4} + C
Substituting back u
I= -\frac{ \cos^{4}x}{4} + C
Final Answer: \boxed{ \int \cos^{3}x\ e^{\log\sin x} \ dx = -\frac{ \cos^{4}x}{4} + C }
Question 16: Integrate \int e^{3\log x}\bigl(x^{4}+1\bigr)^{-1} \ dx
Solution:
Let, I = \int e^{3\log x}\bigl(x^{4}+1\bigr)^{-1} \ dx
We Know, e^{3\log x}=x^{3}\quad(\text{for }x>0)
I=\int \frac{x^{3}}{x^{4}+1},dx
\text{Put }t=x^{4}+1\Rightarrow dt=4x^{3} dx\Rightarrow x^{3} dx=\frac{dt}{4}
I=\frac{1}{4}\int\frac{1}{t} \ dt
I=\frac{1}{4}\log|t|+C
Substituting back t
I =\frac{1}{4}\log\bigl|x^{4}+1\bigr|+C
Final Answer: \boxed{\int e^{3\log x}(x^{4}+1)^{-1} \ dx = \frac{1}{4} \log\bigl|x^{4}+1\bigr|+C}
Question 17: Integrate \int f'(ax+b) \cdot [f(ax+b)]^{n} \ dx
Solution:
Let I = \int f'(ax+b) \cdot [f(ax+b)]^{n} \ dx
\text{Let }u=f(ax+b)\Rightarrow du=f'(ax+b)\cdot adx\Rightarrow f'(ax+b)dx=\frac{du}{a}
I=\int \frac{1}{a}u^{n} \ du
I=\frac{1}{a}\cdot\frac{u^{n+1}}{n+1}+C
Substituting back u
I=\frac{1}{a}\cdot\frac{[f(ax+b)]^{n+1}}{n+1}+C
Final Answer: \boxed{ \int f'(ax+b) [f(ax+b)]^{n} \ dx = \frac{[f(ax+b)]^{n+1}}{a(n+1)}+C}
Question 18: Integrate \int \frac{1}{\sqrt{\sin^{3}x \cdot \sin(x+\alpha)}} \ dx
Solution:
Let, I = \int \frac{1}{\sqrt{\sin^{3}x \cdot \sin(x+\alpha)}} \ dx
I = \int \frac{1}{\sin x\sqrt{\sin x \cdot \sin(x+\alpha)}} \ dx
I = \int \frac{1}{\sin x\sqrt{\sin x [\sin x \cos \alpha + \cos x \sin \alpha]}} \ dx
I = \int \frac{1}{\sin x\sqrt{\sin^{2}x [\cos \alpha + \cot x \sin \alpha]}} \ dx
I = \int \frac{1}{\sin^{2} x\sqrt{\cos \alpha + \cot x \sin \alpha}} \ dx
I = \int \frac{\cosec^{2} x}{\sqrt{\cos \alpha + \cot x \sin \alpha}} \ dx
Let, \cos \alpha + \cot x \sin \alpha = t
-\cosec^{2} x \sin \alpha dx = dt
\cosec^{2} x dx = \frac{-1}{\sin \alpha} dt
I = \frac{-1}{\sin \alpha} \int \frac{1}{\sqrt{t}} \ dx
I = \frac{-2}{\sin \alpha} \sqrt{t} +C
Substituting back t
I = \frac{-2}{\sin \alpha} \int \sqrt{\cos \alpha + \cot x \sin \alpha} + C
I = \frac{-2}{\sin \alpha} \int \sqrt{\cos \alpha + \frac{\cos x}{\sin x} \sin \alpha} + C
I = \frac{-2}{\sin \alpha} \int \sqrt{\frac{\sin (x+ \alpha)}{\sin x}} + C
Final Answer: \boxed{ \int \frac{1}{\sqrt{\sin^{3}x \cdot \sin(x+\alpha)}} \ dx = \frac{-2}{\sin \alpha} \int \sqrt{\frac{\sin (x+ \alpha)}{\sin x}} + C }
Question 19: Integrate \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \ dx
Solution:
Let, I = \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \ dx
Put \sqrt{x} = \cos A \Rightarrow x = \cos^2 A \Rightarrow dx = -2 \cdot \cos A \cdot \sin A \ dA
I = \int \sqrt{\frac{1-\cos A}{1+\cos A}} \ (-2 \cdot \cos A \cdot \sin A dA)
I = \int \sqrt{\frac{2 \sin^2 {\frac{A}{2}}}{2 \cos^2 {\frac{A}{2}}}} \ (-2 \cdot \cos A \cdot \sin A dA)
I = \int \tan{\frac{A}{2}} \ (-2 \cdot \cos A \cdot \sin A dA)
I = \int -2 \cdot \frac{\sin{\frac{A}{2}}}{\cos{\frac{A}{2}}} \cdot \cos A \cdot 2\sin \frac{A}{2} \cdot \cos \frac{A}{2} dA
I = \int -2 \cdot \sin{\frac{A}{2}} \cdot \cos A \cdot 2\sin \frac{A}{2} \ dA
I = \int -4 \cdot \sin^{2}{\frac{A}{2}} \cdot \cos A \ dA
I = \int -4 \cdot \sin^{2}{\frac{A}{2}} \cdot \cos A \ dA
I = \int -2 (1- \cos A) \cdot \cos A \ dA
I = \int -2 (1- \cos A) \cdot \cos A \ dA
I = \int 2 \cos^{2} A - 2 \cos A \ dA
I = \int \cos 2A +1 - 2\cos A \ dA
I = \frac{\sin 2A}{2} + A - 2\sin A + C
I = \frac{2 \sin A \cos A}{2} + A - 2\sin A + C
I = \sin A \cos A + A - 2\sin A} + C
Since: \sqrt{x} = \cos A
Therefore,
I = \sqrt{x}\sqrt{1-x} + \cos^{-1} x - 2\sqrt{1-x} + C
Final Answer: \boxed{\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \ dx = \sqrt{x-x^2} + \cos^{-1} x - 2\sqrt{1-x} + C }
Question 20: Integrate \int \frac{2+\sin 2x}{1+\cos 2x} \ e^{x} \ dx
Solution:
Let I = \int \frac{2+\sin 2x}{1+\cos 2x} \ e^{x} \ dx
Use formula 1+\cos 2x=2\cos^{2}x,\ \sin 2x=2\sin x\cos x
\frac{2+\sin 2x}{1+\cos 2x}=\frac{2+2\sin x\cos x}{2\cos^{2}x}=\frac{1+\sin x\cos x}{\cos^{2}x}=\sec^{2}x+\tan x
I=\int (\sec^{2}x+\tan x)e^{x} \ dx
I=\int e^{x}\sec^{2}x \ dx+\int e^{x}\tan x \ dx
\text{Integrate by parts for the first term: set }u=e^{x} \ dv=\sec^{2}x \ dx\Rightarrow du=e^{x}dx \ v=\tan x
\int e^{x}\sec^{2}x \ dx=e^{x}\tan x-\int e^{x}\tan x \ dx
Substituting in I
I=e^{x}\tan x-\int e^{x}\tan x \ dx+\int e^{x}\tan x \ dx
I=e^{x}\tan x + C
Final Answer: \boxed{\int \frac{2+\sin 2x}{1+\cos 2x} \ e^{x} \ dx=e^{x}\tan x + C}
Question 21: Integrate \int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} \ dx
Solution:
Let, I=\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)},dx
Using Partial fractions:
\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{x+2}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}
x^{2}+x+1=A(x+1)^{2}+B(x+1)(x+2)+C(x+2)
\text{Expanding and comparing coefficients gives }A=3,\ B=-2,\ C=1
\therefore \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{3}{x+2}+\frac{-2}{x+1}+\frac{1}{(x+1)^{2}}
I=3\int\frac{1}{x+2} \ dx-2\int\frac{1}{x+1} \ dx+\int\frac{1}{(x+1)^{2}} \ dx
I=3\log|x+2|-2\log|x+1|-\frac{1}{x+1}+C
Final Answer: \boxed{\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} \ dx=3\ln|x+2|-2\ln|x+1|-\frac{1}{x+1}+C}
Question 22: Integrate \int \tan^{-1} \sqrt{\frac{1-x}{1+x}} \ dx
Solution:
Let, I = \int \tan^{-1} \sqrt{\frac{1-x}{1+x}} \ dx
\text{Put }x=\cos 2\theta\Rightarrow \sqrt{\frac{1-x}{1+x}}=\sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}}=\tan\theta
I=\int \tan^{1} (\tan\theta)dx
I=\int \theta dx
\text{Since: }x=\cos 2\theta \Rightarrow \theta = \frac{1}{2} \cos^{-1} x
I=\int \frac{1}{2} \cos^{-1} x dx
I= \frac{1}{2} \int 1 \cdot \cos^{-1} x dx
Applying by-parts
u = \cos^{-1} x \rightarrow du = - \frac{1}{\sqrt{1-x^2}}, \quad dv = 1 \rightarrow v = x
I= \frac{1}{2} \left[ x\cos^{-1} x - \int x \cdot \frac{-1}{\sqrt{1-x^2}} \ dx \right]
I= \frac{1}{2} \left[ x\cos^{-1} x + \int x \cdot \frac{1}{\sqrt{1-x^2}} \ dx \right]
For second integral let, 1-x^2 = t \rightarrow -2x dx = dt \rightarrow xdx = -\frac{1}{2} dt
I= \frac{1}{2} \left[ x\cos^{-1} x - \int \frac{1}{2} \cdot \frac{1}{\sqrt{t}} \ dt \right]
I= \frac{1}{2} \left[ x\cos^{-1} x - \frac{1}{2} \cdot \frac{2}{1} \cdot {\sqrt{t}} \right] + C
I= \frac{1}{2} \left[ x\cos^{-1} x + {\sqrt{t}} \right] + C
I= \frac{1}{2} \left[ x\cos^{-1} x + {\sqrt{1 - x^2}} \right] + C
Final Answer: \boxed{ \int \tan^{-1} \sqrt{\frac{1-x}{1+x}} \ dx= \frac{1}{2} \left[ x\cos^{-1} x + {\sqrt{1 - x^2}} \right] + C }
Question 23: Integrate \int \frac{\sqrt{x^{2}+1}\,\bigl[\log(x^{2}+1)-2\log x\bigr]}{x^{4}} \ dx
Solution:
Let, I=\int \frac{\sqrt{x^{2}+1} \bigl(\log(x^{2}+1)-2\log x\bigr)}{x^{4}} \ dx
\text{Note }\log(x^{2}+1)-2\log x=\log\frac{x^{2}+1}{x^{2}}
I=\int \frac{\sqrt{x^{2}+1} \log \left| \frac{x^{2}+1}{x^{2}} \right| }{x^{4}} \ dx
\text{Put }u=\frac{1}{x} \Rightarrow x=\frac{1}{u} \implies dx=-\frac{1}{u^{2}} \ du
\frac{x^{2}+1}{x^2}=1+u^{2},\ \sqrt{x^{2}+1}=\frac{\sqrt{1+u^{2}}}{u},\ x^{4}=\frac{1}{u^{4}}
I=-u\sqrt{1+u^{2}}\ln(1+u^{2}) \ du
I=-\int u\sqrt{1+u^{2}}\ln(1+u^{2}) \ du
\text{Put }v=1+u^{2}\Rightarrow dv=2u,du\Rightarrow u,du=\frac{dv}{2}
I=-\frac{1}{2}\int v^{1/2}\ln v \ dv
\text{Integrate by parts: }p=\ln v,\ dq=v^{1/2}dv\Rightarrow dp=\frac{dv}{v},\ q=\tfrac{2}{3}v^{3/2}
I=-\frac{1}{2}\Big(\frac{2}{3}v^{3/2}\ln v-\int \frac{2}{3}v^{3/2}\cdot\frac{1}{v} \ dv\Big)
=-\frac{1}{3}v^{3/2}\log v+\frac{1}{3}\cdot\frac{2}{3}v^{3/2}+C=-\frac{1}{3}v^{3/2}\Big(\log v-\frac{2}{3}\Big)+C
\text{Back-substitute }v=1+u^{2}=1+\tfrac{1}{x^{2}}\Rightarrow v^{3/2}=\frac{(x^{2}+1)^{3/2}}{x^{3}}
I=-\frac{1}{3} \frac{(x^{2}+1)^{3/2}}{x^{3}}\Big(\log\bigl(1+\tfrac{1}{x^{2}}\bigr)-\frac{2}{3}\Big)+C
Final Answer: \boxed{ \int \frac{\sqrt{x^{2}+1} \bigl(\log(x^{2}+1)-2\log x\bigr)}{x^{4}} \ dx=-\frac{1}{3} \left( \frac{x^{2}+1}{x^{2}} \right)^{3/2}\Big(\log\bigl(1+\frac{1}{x^{2}}\bigr)-\frac{2}{3}\Big)+C }
Question 24: Integrate \int_{\pi/2}^{\pi} e^{x} \ \left( \frac{1-\sin x}{1-\cos x} \right) \ dx
Solution:
Let, I=\int_{\pi/2}^{\pi} e^{x} \left( \frac{1-\sin x}{1-\cos x} \right) dx
I=\int_{\pi/2}^{\pi} e^{x} \left( \frac{1-2\sin \frac{x}{2}\cos\frac{x}{2}}{2\sin^{2} \frac{x}{2}} \right) dx
I=\int_{\pi/2}^{\pi} e^{x} \left( \frac{1}{{2\sin^{2} \frac{x}{2}}} - \frac{2\sin \frac{x}{2}\cos\frac{x}{2}}{2\sin^{2} \frac{x}{2}} \right) dx
I=\int_{\pi/2}^{\pi} e^{x} \left( \frac{1}{2}\cosec^{2} \frac{x}{2} - \cot \frac{x}{2} \right) dx
I=\int_{\pi/2}^{\pi} e^{x} \left( - \cot \frac{x}{2} + \frac{1}{2}\cosec^{2} \frac{x}{2} \right) dx
We know \int e^x (f(x) + f'(x)) \ dx = e^x f(x) + C
Therefore;
I=\left[- e^{x} \cot \frac{x}{2} \right]_{\pi/2}^{\pi}
I=\left[e^{x} \cot \frac{x}{2} \right]_{\pi}^{\pi/2}
I= (e^{\pi/2} \cot \frac{\pi}{4}) - (e^{\pi} \cot \frac{\pi}{2})
I= e^{\frac{\pi}{2}}
Final Answer: \boxed{ \int_{\pi/2}^{\pi} e^{x} \left( \frac{1-\sin x}{1-\cos x} \right) dx = e^{\frac{\pi}{2}} }
Question 25: Integrate \int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^{4}x+\sin^{4}x} \ dx
Solution:
Let, I=\int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^{4}x+\sin^{4}x} \ dx
Dividing numerator and denominator by \cos^{4} x
I=\int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^{4}x+\sin^{4}x} \ dx
I=\int_{0}^{\pi/4}\frac{\tan x\sec^2 x}{1+\tan^{4}x} \ dx
\text{Let }t =\tan^{2} x \Rightarrow dt= 2\tan x \sec^{2} x dx \Rightarrow \frac{dt}{2}= \tan x \sec^{2} x dx
x = 0 \Rightarrow t = 0, \quad x = \frac{\pi}{4} \Rightarrow t = 1
I= \frac{1}{2} \int_{0}^{1}\frac{1}{1+t^{2}} \ dt
I = \frac{1}{2} \left[ tan^{-1} t \right]_{0}^{1} + C
I = \frac{1}{2} \left[ \frac{\pi}{4} - 0 \right]_{0}^{1} + C
I = \frac{\pi}{8} + C
Final Answer: \boxed{\int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^{4}x+\sin^{4}x} \ dx = \frac{\pi}{8} + C }
Question 26: Evaluate \int_{0}^{\pi/2}\frac{\cos^{2}x}{\cos^{2}x+4\sin^{2}x} \ dx
Solution:
Let, I = \int_{0}^{\pi/2}\frac{\cos^{2}x}{\cos^{2}x+4\sin^{2}x} \ dx
dividing numerator and denominator by \cos^{4} x
I=\int_{0}^{\pi/2}\frac{\sec^{2}x}{\sec^{2}x+4\sec^{2}x \tan^{2} x} \ dx
Let \tan x = t \Rightarrow \sec^{2} x dx = dt
x = 0 \Rightarrow t = 0, \quad x = \frac{\pi}{2} \Rightarrow t \rightarrow \infty
I=\int_{0}^{\infty}\frac{dt}{\sec^{2}x +4\sec^{2}x \cdot t^2} \ dx
I=\int_{0}^{\infty}\frac{dt}{\sec^{2}x (1 + 4 t^2)} \ dx
I=\int_{0}^{\infty}\frac{dt}{ (1 + \tan^{2}x) (1 + 4 t^2)} \ dx
I=\int_{0}^{\infty}\frac{dt}{ (1 + t^{2}) (1 + 4 t^2)} \ dx
Using Partial Fraction
\frac{dt}{ (1 + t^{2}) (1 + 4 t^2)} = \frac{Ax+B}{(1 + t^{2})} + \frac{Cx+D}{(1 + 4 t^2)}
1 = (Ax+B)(1 + 4 t^2) + (Cx+D)(1 + t^{2})
1 = At + 4At^3 + B + 4Bt^2 + Ct + Ct^3 + D + Dt^2
1 = t^3(4A+C) + t^2(4B+D) + t(A+C) + (B + D)
Comparing coefficients of
t^3: 4A + C = 0
t^2: 4B + D = 0
t: A + C = 0
constants: B + D = 1
Solving the above equation: A = 0, B = -\frac{1}{3}, C = 0, D = \frac{4}{3}
\frac{dt}{ (1 + t^{2}) (1 + 4 t^2)} = \frac{-\frac{1}{3}}{(1 + t^{2})} + \frac{\frac{4}{3}}{(1 + 4 t^2)}
I = -\frac{1}{3} \int_{0}^{\infty} \frac{dt}{(1 + t^{2})} + \frac{4}{3} \int \frac{dt}{(1 + (2t)^2)}
I = -\frac{1}{3} \left[ \tan^{-1} t + \frac{4}{3} \cdot \frac{1}{2} \tan^{-1} (2t) \right]_{0}^{\infty}
I = \frac{1}{3} \left[ -\tan^{-1} t + 2 \tan^{-1} (2t) \right]_{0}^{\infty}
I = \frac{1}{3} \left[ 2 \tan^{-1} (2t) - \tan^{-1} t \right]_{0}^{\infty}
I = \frac{1}{3} [ (2 \frac{\pi}{2}) - \frac{\pi}{2}) - (0) ]
I = \frac{1}{3} [ \frac{\pi}{2}]
I = \frac{\pi}{6}
Final Answer: \int_{0}^{\pi/2}\frac{\cos^{2}x}{\cos^{2}x+4\sin^{2}x} \ dx = \frac{\pi}{6}
Question 27: Evaluate \int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin 2x}} \ dx
Solution:
Let, I = \int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin 2x}} \ dx
\text{Put }u=\sin x-\cos x,\quad du=(\cos x+\sin x)dx
(\sin x-\cos x)^2 = \sin^{2} x + \cos^{2} x - 2\sin x \cos x
(\sin x-\cos x)^2 = 1 - \sin {2x}
u^2 = 1 - \sin 2x
\sin 2x=1-u^{2},\ \sqrt{\sin 2x}=\sqrt{1-u^{2}}
I=\int\frac{du}{\sqrt{1-u^{2}}} du
I= \sin^{-1} u
Substituting back u:
I= \left[ \sin^{-1} (\sin x-\cos x) \right]_{\pi/6}^{\pi/3}
I=\sin^{-1}\Big(\tfrac{\sqrt{3}-1}{2}\Big)-\sin^{-1}\Big(\tfrac{1-\sqrt{3}}{2}\Big)
I=\sin^{-1}\Big(\tfrac{\sqrt{3}-1}{2}\Big)+\sin^{-1}\Big(\tfrac{\sqrt{3}-1}{2}\Big)
I=2\sin^{-1}\Big(\tfrac{\sqrt{3}-1}{2}\Big)
Final Answer: \boxed{\int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin 2x}} \ dx = 2\sin^{-1} \Big(\tfrac{\sqrt{3}-1}{2}\Big)}
Question 28: Evaluate \int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}
Solution:
I = \int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}
Rationalizing the denominator
\frac{1}{\sqrt{1+x}-\sqrt{x}} = \frac{1}{\sqrt{1+x}-\sqrt{x}} \cdot \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}}
\frac{1}{\sqrt{1+x}-\sqrt{x}} = \frac{\sqrt{1+x}-\sqrt{x}}{1+x-x}
\frac{1}{\sqrt{1+x}-\sqrt{x}} = \frac{\sqrt{1+x}-\sqrt{x}}{1}
I=\int_{0}^{1}(\sqrt{1+x}+\sqrt{x}) \ dx
I=\frac{2}{3}\big[(1+x)^{3/2}\big]_{0}^{1}+\frac{2}{3} \big[ x^{3/2} \big]_{0}^{1}
I=\frac{2}{3}(2\sqrt{2}-1)+\frac{2}{3}
I=\frac{4\sqrt{2}}{3}
Final Answer: \boxed{ \int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}} = \frac{4\sqrt{2}}{3}}
Question 29: Evaluate: \int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x} \ dx
Solution:
Let, I = \int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x} \ dx
\text{Put }u=\sin x-\cos x,\quad du=(\cos x+\sin x)dx
u^2 = (\sin x - \cos x)^2
u^2 = \sin^2x + \cos^2x - 2 \sin x \cos x
u^2 = 1 - \sin 2x
\sin 2x=1-u^{2}
When x = 0, u = -1, \quad x = \frac{\pi}{4}, u = 0
I=\int_{-1}^{0}\frac{du}{9+16(1-u^{2})}
I=\int_{-1}^{0}\frac{du}{25-16u^{2}}
I= \frac{1}{16} \int_{-1}^{0} \frac{du}{ \frac{25}{16}-u^{2}}
I= \frac{1}{16} \int_{-1}^{0} \frac{du}{ (\frac{5}{4})^2-u^{2}}
Using formula: \int\frac{du}{a^{2}-u^{2}}=\frac{1}{2a}\log \Big|\frac{a+u}{a-u}\Big|
Here a=\frac{5}{4}.
I= \frac{1}{16} \frac{2}{5} \log \Big|\frac{5+4u}{5-4u}\Big|\Big|_{-1}^{0}
I=\frac{1}{40} \big[ \log \Big|\frac{5}{5}\Big| - \log \Big|\frac{1}{9}\Big| \big]
I= \frac{1}{40} \log 9
Final Answer: \boxed{ \int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x} \ dx = \frac{1}{40} \log 9 }
Question 30: Evaluate \int_{0}^{\pi/2}\sin 2x\cdot \tan^{-1}(\sin x) \ dx
Solution:
Let I = \int_{0}^{\pi/2}\sin 2x\cdot \tan^{-1}(\sin x) \ dx
Use integration by parts:
u=\tan^{-1}(\sin x), \ dv=\sin 2x,dx
du=\frac{\cos x}{1+\sin^{2}x}dx,\quad v=-\frac{1}{2}\cos 2x
I=\Big[-\frac{1}{2}\cos 2x\cdot \tan^{-1}(\sin x)\Big]_{0}^{\pi/2}+\frac{1}{2}\int_{0}^{\pi/2}\cos 2x\cdot\frac{\cos x}{1+\sin^{2}x} \ dx
I=\Big[-\frac{1}{2}\cos 2x\cdot \tan^{-1}(\sin x)\Big]_{0}^{\pi/2}+\frac{1}{2}\int_{0}^{\pi/2}(1-2\sin^{2}x)\cdot\frac{\cos x}{1+\sin^{2}x} \ dx
For \int (1-2\sin^{2}x)\cdot\frac{\cos x}{1+\sin^{2}x} \ dx
Remaining integral: put t=\sin x, \quad dt = \cos x dx :
\frac{1}{2} \int (1-2\sin^{2}x)\cdot\frac{\cos x}{1+\sin^{2}x} \ dx = \frac{1}{2}\int \frac{1-2t^{2}}{1+t^{2}} \ dt
\frac{1}{2} \int (1-2\sin^{2}x)\cdot\frac{\cos x}{1+\sin^{2}x} \ dx = -\frac{1}{2}\int \frac{2t^{2}-1}{t^{2}+1} \ dt
\frac{1}{2} \int (1-2\sin^{2}x)\cdot\frac{\cos x}{1+\sin^{2}x} \ dx = -\frac{1}{2}\int \frac{2t^{2} +2 - 2 -1}{t^{2}+1} \ dt
\frac{1}{2} \int (1-2\sin^{2}x)\cdot\frac{\cos x}{1+\sin^{2}x} \ dx = -\frac{1}{2}\int \frac{2t^{2} +2 - 3}{t^{2}+1} \ dt
\frac{1}{2} \int (1-2\sin^{2}x)\cdot\frac{\cos x}{1+\sin^{2}x} \ dx = -\frac{1}{2}\int \frac{2t^{2} +2}{t^{2}+1} - \frac{3}{t^{2}+1} \ dt
\frac{1}{2} \int (1-2\sin^{2}x)\cdot\frac{\cos x}{1+\sin^{2}x} \ dx = -\frac{1}{2}\int \frac{2(t^{2} +1)}{t^{2}+1} - \frac{3}{t^{2}+1} \ dt
\frac{1}{2} \int (1-2\sin^{2}x)\cdot\frac{\cos x}{1+\sin^{2}x} \ dx = -\frac{1}{2}\int 2 - \frac{3}{t^{2}+1} \ dt
\frac{1}{2} \int (1-2\sin^{2}x)\cdot\frac{\cos x}{1+\sin^{2}x} \ dx = -\frac{1}{2}\int 2 - \frac{3}{t^{2}+1} \ dt
\frac{1}{2} \int (1-2\sin^{2}x)\cdot\frac{\cos x}{1+\sin^{2}x} \ dx = - t + \frac{3}{2} \tan^{-1}{t}
Substituting back t;
\int (1-2\sin^{2}x)\cdot\frac{\cos x}{1+\sin^{2}x} \ dx = - \sin x + \frac{3}{2} \tan^{-1}{\sin x}
Substituting back in I
I=\left[-\frac{1}{2}\cos 2x\cdot \tan^{-1}(\sin x) + \frac{3}{2} \tan^{-1}{\sin x} - \sin x \right]_{0}^{\pi/2}
I=\left[-\frac{1}{2}\cos \pi\cdot \tan^{-1}(\sin \frac{\pi}{2}) + \frac{3}{2} \tan^{-1}{\sin \frac{\pi}{2}} - \sin \frac{\pi}{2} \right] - \left[-\frac{1}{2}\cos 0 \cdot \tan^{-1}(\sin 0) + \frac{3}{2} \tan^{-1}{\sin 0} - \sin 0 \right]
I=\left[ -\frac{1}{2} (-1) \cdot \frac{\pi}{4} + \frac{3}{2} \cdot \frac{\pi}{4} - 1 \right] - \left[-\frac{1}{2} \cdot 1 \cdot 0 + \frac{3}{2} \cdot 0 - 0 \right]
I=\left[\frac{\pi}{8} + \frac{3\pi}{8} - 1 \right] - \left[0 + 0 + 0 \right]
I=\left[\frac{4\pi}{8} - 1 \right]
I=\left[\frac{\pi}{2} - 1 \right]
Final Answer: \boxed{\int_{0}^{\pi/2}\sin 2x\cdot \tan^{-1}(\sin x) \ dx = \frac{\pi}{2}-1}
Question 31: Evaluate: \int_{1}^{4} \bigl( |x-1|+|x-2|+|x-3| \bigr) \ dx
Solution:
Let I = \int_{1}^{4} \bigl( |x-1|+|x-2|+|x-3| \bigr) \ dx
\text{Split interval at }x=2,3;
\text{For }1\le x\le2:\ |x-1|=x-1,\ |x-2|=2-x,\ |x-3|=3-x
\Rightarrow \bigl( |x-1|+|x-2|+|x-3| \bigr)=4-x
\text{For }2\le x\le3:\ |x-1|=x-1,\ |x-2|=x-2,\ |x-3|=3-x
\Rightarrow \bigl( |x-1|+|x-2|+|x-3| \bigr)=x
\text{For }3\le x\le4:\ |x-1|=x-1,\ |x-2|=x-2,\ |x-3|=x-3
\Rightarrow \bigl( |x-1|+|x-2|+|x-3| \bigr)=3x-6
I=\int_{1}^{2}(4-x) \ dx+\int_{2}^{3}x \ dx+\int_{3}^{4}(3x-6) \ dx
I= \left[ 4x-\frac{x^{2}}{2} \right]_{1}^{2}+ \left[ \frac{x^{2}}{2} \right]_{2}^{3}+ \left[ \frac{3x^{2}}{2}-6x \right]_{3}^{4}
I=\frac{5}{2}+\frac{5}{2}+\frac{9}{2}
I = \frac{19}{2}
Final Answer: \boxed{ I = \int_{1}^{4} \bigl( |x-1|+|x-2|+|x-3| \bigr) \ dx = \frac{19}{2}}
Question 32: Prove that \int_{1}^{3}\frac{dx}{x^{2}(x+1)} = \frac{2}{3}+\ln\frac{2}{3}
Solution:
LHS = \int_{1}^{3}\frac{dx}{x^{2}(x+1)}
Using Partial Fractions:
\frac{1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}
1=A x(x+1)+B(x+1)+C x^{2}
1=(A+C)x^{2}+(A+B)x+B
Comparing coefficients:
A+C=0,\ A+B=0,\ B=1\Rightarrow A=-1,\ B=1,\ C=1
Therefore,
\frac{1}{x^{2}(x+1)}=\frac{-1}{x}+\frac{1}{x^{2}}+\frac{1}{x+1}
LHS=\int_{1}^{3}\Big(-\frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{x+1}\Big)dx
LHS=\Big(-\ln x-\frac{1}{x}+\ln(x+1)\Big)_{1}^{3}
LHS=\Big(\ln\frac{4}{3}-\frac{1}{3}\Big)-\Big(\ln 2-1\Big)
LHS=\ln\frac{2}{3}+\frac{2}{3}
\boxed{\displaystyle \int_{1}^{3}\frac{dx}{x^{2}(x+1)}=\frac{2}{3}+\ln\frac{2}{3}} Hence Proved.
Question 33: Prove that \int_{0}^{1} x e^{x} \ dx = 1
Solution:
LHS \int_{0}^{1} x e^{x} \ dx
\text{Integration by parts: }u=x,\ dv=e^{x}dx\Rightarrow du=dx,\ v=e^{x}
LHS=\Big[x e^{x}\Big]_{0}^{1}-\int_{0}^{1}e^{x} \ dx
LHS=\Big[x e^{x}\Big]_{0}^{1}- \Big[e^{x} \Big]_{0}^{1}
LHS=e\ -\ (e-1)=1
\boxed{\displaystyle \int_{0}^{1}x e^{x} \ dx=1} Hence Proved
Question 34: Prove that \int_{-1}^{1} x^{17}\cos^{4}x \ dx = 0
Solution:
LHS = \int_{-1}^{1} x^{17}\cos^{4}x \ dx
x^{17}\text{ is odd, }\cos^{4}x\text{ is even}\Rightarrow x^{17}\cos^{4}x\text{ is odd}
\text{By property }P_{7}\text{ (odd function over }[-a,a]\text{): integral }=0
Therefore:
LHS = 0
\boxed{\displaystyle \int_{-1}^{1}x^{17}\cos^{4}x \ dx=0} Hence Proved
Question 35: Prove that \int_{0}^{\pi/2}\sin^{3}x \ dx = \frac{2}{3}
Solution:
LHS = \int_{0}^{\pi/2}\sin^{3}x \ dx
\sin^{3}x=\sin x(1-\cos^{2}x)
LHS= \int_{0}^{\pi/2}\sin x \ dx-\int_{0}^{\pi/2}\sin x\cos^{2}x \ dx
\int_{0}^{\pi/2}\sin x \ dx=[- \cos x]_{0}^{\pi/2}=1
\text{Put }u=\cos x \ du=-\sin x \ dx\Rightarrow \int_{0}^{\pi/2}\sin x\cos^{2}x \ dx=\int_{1}^{0}-u^{2} \ du=\int_{0}^{1}u^{2} \ du=\frac{1}{3}
Substituting in LHS
LHS=1-\tfrac{1}{3}=\tfrac{2}{3}
\boxed{ \int_{0}^{\pi/2}\sin^{3}x,dx=\tfrac{2}{3}} Hence Proved
Question 36: Prove that: \int_{0}^{\pi/4} 2\tan^{3}x \ dx = 1-\ln 2
Solution:
LHS = \int_{0}^{\pi/4} 2\tan^{3}x \ dx
\tan^{3}x=\tan x \cdot \tan^{2}x = \tan x(\sec^{2}x-1)=\tan x\sec^{2}x-\tan x
LHS=2\int_{0}^{\pi/4}(\tan x\sec^{2}x) \ dx-2\int_{0}^{\pi/4} \tan x \ dx
\text{Put }u=\tan x\Rightarrow du=\sec^{2}x \ dx\Rightarrow 2\int \tan x\sec^{2}x \ dx=2\cdot\frac{u^{2}}{2}=\tan^{2}x
\int \tan x \ dx=-\log|\cos x|\Rightarrow -2\int\tan x \ dx=2\log|\cos x|
LHS = \left[ \tan^{2}x+2\log|\cos x| \right]_{0}^{\pi/4}
LHS=\big(1+2\log\frac{1}{\sqrt{2}}\big)-\big(0+2\log 1\big)
LHS =1- \log 2
\boxed{\int_{0}^{\pi/4} 2\tan^{3}x \ dx=1-\log 2} Hence Proved
Question 37: Prove that \int_{0}^{1} \sin^{-1}x \ dx = \frac{\pi}{2}-1
Solution:
LHS = \int_{0}^{1}\sin^{-1}x \ dx
\text{Integration by parts: }u=\sin^{-1}x,\ dv=dx \Rightarrow du=\frac{dx}{\sqrt{1-x^{2}}}, \ v=x
LHS =\Big[x\sin^{-1}x\Big]_{0}^{1}-\int_{0}^{1}\frac{x}{\sqrt{1-x^{2}}} \ dx
\Big[x\sin^{-1}x\Big]_{0}^{1} = 1\cdot\frac{\pi}{2}-0=\frac{\pi}{2}
\text{Put }t=1-x^{2}\Rightarrow dt=-2x,dx\Rightarrow \int_{0}^{1}\frac{x}{\sqrt{1-x^{2}}},dx=1
LHS =\frac{\pi}{2}-1
\boxed{ \int_{0}^{1}\sin^{-1}x,dx=\frac{\pi}{2}-1} Hence Proved
Question 38: Integrate \int \frac{dx}{e^{x}+e^{-x}}
\text{(A)}\ \tan^{-1}(e^{x}) + C
\text{(B)}\ \tan^{-1}(e^{x}) + C
\text{(C)}\ \log(e^{x} - e^{-x}) + C
\text{(D)}\ \log(e^{x} + e^{-x}) + C
Solution:
Let, I = \int \frac{dx}{e^{x}+e^{-x}}
I=\int \frac{e^{x},dx}{e^{2x}+1}
\text{Put }t=e^{x}\Rightarrow dt=e^{x}dx
I = \int\frac{dt}{t^{2}+1}
I =\tan^{-1}t+C
I =\tan^{-1}(e^{x})+C
Correct option: (A)
Question 39: Evaluate \int \frac{\cos 2x}{(\sin x+\cos x)^{2}} \ dx
\text{(A)}\ -\frac{1}{\sin x + \cos x} + C
\text{(B)}\ \log|\sin x + \cos x| + C
\text{(C)}\ \log|\sin x - \cos x| + C
\text{(D)}\ \frac{1}{(\sin x + \cos x)^{2}}
Solution:
Let, I = \int \frac{\cos 2x}{(\sin x+\cos x)^{2}} \ dx
\cos2x=\cos^{2}x-\sin^{2}x=(\cos x-\sin x)(\cos x+\sin x)
\frac{\cos2x}{(\sin x+\cos x)^{2}}=\frac{(\cos x-\sin x)(\sin x+\cos x)}{(\sin x+\cos x)^{2}}=\frac{\cos x-\sin x}{\sin x+\cos x}
I = \int \frac{\cos x-\sin x}{\sin x+\cos x} \ dx
\text{Put }u=\sin x+\cos x \Rightarrow du=(\cos x-\sin x) \ dx
I = \int\frac{du}{u}
I =\ln|u|+C
I =\ln|\sin x+\cos x|+C
Correct option: (B)
Question 40: If f(a+b-x)=f(x) , evaluate \int_{a}^{b} x f(x) \ dx
\text{(A)}\ \frac{a+b}{2}\int_{a}^{b} f(b-x) \ dx
\text{(B)}\ \frac{a+b}{2}\int_{a}^{b} f(b+x) \ dx
\text{(C)}\ \frac{b-a}{2}\int_{a}^{b} f(x) \ dx
\text{(D)}\ \frac{a+b}{2}\int_{a}^{b} f(x) \ dx
Solution:
Let, I=\int_{a}^{b}x f(x) \ dx … (1)
Using Property: \int_{a}^{b} f(x) \ dx = \int_{a}^{b} f(a+b-x) \ dx
I=\int_{a}^{b} (a+b-x) f(a+b-x) \ dx
given f(a+b-x)=f(x)
I=\int_{a}^{b}(a+b-x)f(x) \ dx … (2)
Adding (1) and (2)
2I =\int_{a}^{b}\big(x+(a+b-x)\big)f(x) \ dx
2I =(a+b)\int_{a}^{b}f(x) \ dx
I =\frac{a+b}{2}\int_{a}^{b}f(x) \ dx
Correct option: (D)