For those of you looking for NCERT solutions for class 12th maths Ex 7.2 , you have come to the right place.
In this post, we provide NCERT Solutions for Class 12 Maths Exercise 7.2 Integration by Substitution. Each solution is explained step by step so students can learn how to apply substitution effectively. These solutions will help you in CBSE board exam preparation and competitive exams.
Question 1: \int \frac{2x}{1+x^2} \ dx
Solution:
Let, I = \int \frac{2x}{1+x^2} \ dx
\text{Let } t = 1 + x^2 \implies dt = 2x \ dx \
I = \int \frac{dt}{t}
I = \log|t| + C \
I = \log(1+x^2) + C \
Final Answer: \boxed{\int \frac{2x}{1+x^2} \ dx = \log(1+x^2) + C}
Question 2: \int \frac{(\log x)^2}{x} \ dx
Solution:
Let, I = \int \frac{(\log x)^2}{x} \ dx
\text{Let } t = \log x \implies dt = \frac{dx}{x} \
I = \int t^2 \ dt
I = \frac{t^3}{3} + C \
I = \frac{(\log x)^3}{3} + C \
Final Answer: \boxed{\int \frac{(\log x)^2}{x} \ dx = \frac{(\log x)^3}{3} + C}
Question 3: \int \frac{1}{x + x\log x} \ dx
Solution:
Let, I = \int \frac{1}{x + x\log x} \ dx
I = \int \frac{1}{x(1+\log x)},dx \
\text{Let } t = 1+\log x \implies dt = \frac{dx}{x} \
I= \int \frac{dt}{t}
I = \log|t| + C \
I = \log(1+\log x) + C \
Final Answer: \boxed{\int \frac{1}{x + x\log x} \ dx = \log(1+\log x) + C}
Question 4: \int \sin x \cdot \sin(\cos x) \ dx
Solution:
Let, I = \int \sin x \cdot \sin(\cos x) \ dx
\text{Let } t = \cos x \implies dt = -\sin x,dx \
I = -\int \sin t,dt \
I = \cos t + C
I = \cos(\cos x) + C \
Final Answer: \boxed{\int \sin x \cdot \sin(\cos x) \ dx = \cos(\cos x) + C}
Question 5: \int \sin(ax+b)\cos(ax+b) \ dx
Solution:
Let, I = \int \sin(ax+b)\cos(ax+b) \ dx
\sin(ax+b)\cos(ax+b) = \frac{2}{2} \sin(ax+b)\cos(ax+b) = \frac{1}{2}\sin(2ax+2b)
I = \tfrac{1}{2}\int \sin(2ax+2b) \ dx
I = \tfrac{1}{2}\cdot \left(\frac{-1}{2a}\cos(2ax+2b)\right) + C
I = -\tfrac{1}{4a}\cos(2ax+2b) + C \
Final Answer: \boxed{\int \sin(ax+b)\cos(ax+b) \ dx = -\tfrac{1}{4a}\cos(2ax+2b) + C}
Question 6: \int \sqrt{ax+b} \ dx
Solution:
Let, I = \int \sqrt{ax+b} \ dx
\text{Let } t = ax+b \implies dt = a \cdot dx \
I = \frac{1}{a}\int t^{1/2} \ dt
I = \frac{1}{a}\cdot \frac{2}{3}t^{3/2} + C
I = \frac{2}{3a}(ax+b)^{3/2} + C
Final Answer: \boxed{\int \sqrt{ax+b} \ dx = \tfrac{2}{3a}(ax+b)^{3/2} + C}
Question 7: \int x \sqrt{x+2} \ dx
Solution:
Let, I = \int x \sqrt{x+2} \ dx
Let, x+2 = t \implies dx = dt
I = \int {t-2} \sqrt{t} \ dt
I = \int t^{3/2} - 2t^{1/2} \ dt
I = \frac{t^{5/2}}{\frac{5}{2}} - 2 \frac{t^{3/2}}{\frac{3}{2}} + C
I = \frac{2t^{5/2}}{5} - \frac{4t^{3/2}}{3} + C
I = \frac{2(x+2)^{5/2}}{5} - \frac{4(x+2)^{3/2}}{3} + C
Final Answer: \boxed{\int x \sqrt{x+2} \ dx =\frac{2}{5}(x+2)^{5/2} - \frac{4}{3}(x+2)^{3/2} + C}
Question 8: \int x\sqrt{1+2x^2} \ dx
Solution:
Let, I = \int x\sqrt{1+2x^2} \ dx
\text{Let } t = 1 + 2x^2 \implies dt = 4x \ dx
I = \frac{1}{4}\int \sqrt{t} \ dt
I = \frac{1}{4}\cdot \frac{2}{3}t^{3/2} + C \
I = \frac{1}{6}(1+2x^2)^{3/2} + C \
Final Answer: \boxed{\int x\sqrt{1+2x^2} \ dx = \frac{1}{6}(1+2x^2)^{3/2} + C}
Question 9: \int (4x+2)\sqrt{x^2+x+1} \ dx
Solution:
I = \int (4x+2)\sqrt{x^2+x+1} \ dx
\text{Let } t = x^2+x+1 \implies dt = (2x+1) dx \
I = 2\int (2x+1)\sqrt{t} \ dx \
I = 2\int \sqrt{t} \ dt
I = 2 \cdot \frac{2}{3}t^{3/2} + C \
I = \frac{4}{3}(x^2+x+1)^{3/2} + C \
Final Answer: \boxed{\int (4x+2)\sqrt{x^2+x+1} \ dx = \frac{4}{3}(x^2+x+1)^{3/2} + C}
Question 10: \int \frac{1}{x-\sqrt{x}} \ dx
Solution
I = \int \frac{1}{x-\sqrt{x}} \ dx
\text{Let } t = \sqrt{x} \implies x = t^2 \implies dx = 2tdt
I = \int \frac{2t}{t^2 - t} \ dt
I = \int \frac{2}{t-1} \ dt
I = 2\log|t-1| + C
I = 2\log|\sqrt{x}-1| + C
Final Answer: \boxed{\int \frac{1}{x-\sqrt{x}} \ dx = 2\log|\sqrt{x}-1| + C}
Question 11: \int \frac{x}{\sqrt{x+4}} \ dx,\quad x>0
Solution:
Let, I = \int \frac{x}{\sqrt{x+4}} \ dx
\text{Put } t = x+4\implies dt = dx \
\text{Then } x = t-4 \
I = \int \frac{t-4}{\sqrt{t}} \ dt
I = \int \bigl(t^{1/2} - 4t^{-1/2}\bigr) \ dt
I = \frac{2}{3}t^{3/2} - 8t^{1/2} + C
I = \frac{2}{3}(x+4)^{3/2} - 8(x+4)^{1/2} + C
I = \frac{2}{3} \sqrt{x+4} \big[(x+4) - 8 \frac{3}{2} \big] + C
I = \frac{2}{3} \sqrt{x+4} \big[(x+4) - 12 \big] + C
I = \frac{2}{3} \sqrt{x+4} \big[x- 8 \big] + C
Final Answer: \boxed{\int \frac{x}{\sqrt{x+4}} \ dx = \frac{2}{3} \sqrt{x+4} \big[x- 8 \big] + C}
Question 12: \int x^{5} \ (x^{3}-1)^{\frac{1}{3}} \ dx
Solution:
Let, I = \int x^{5} \ (x^{3}-1)^{\frac{1}{3}} \ dx
\text{Put } t = x^{3}-1 \implies dt = 3x^{2} \ dx \quad \Rightarrow \quad x^{2} \ dx = \frac{dt}{3}
\text{Write } x^{5} \ dx = x^{3} \cdot x^{2} \ dx = (t+1) \cdot \frac{dt}{3}
I = \frac{1}{3}\int (t+1)t^{1/3} \ dt
I = \frac{1}{3}\int \bigl(t^{4/3} + t^{1/3}\bigr) \ dt
I = \frac{1}{3}\Bigl(\frac{3}{7}t^{7/3} + \frac{3}{4}t^{4/3}\Bigr) + C
I = \frac{1}{7}t^{7/3} + \frac{1}{4}t^{4/3} + C
I = \frac{1}{7}(x^{3}-1)^{7/3} + \frac{1}{4}(x^{3}-1)^{4/3} + C
Final Answer: \boxed{\int x^{5} \ (x^{3}-1)^{\frac{1}{3}} \ dx = \frac{1}{7}(x^{3}-1)^{7/3} + \frac{1}{4}(x^{3}-1)^{4/3} + C}
Question 13: \int \frac{x^{2}}{\bigl(2+3x^{3}\bigr)^{3}} \ dx
Solution:
Let I = \int \frac{x^{2}}{\bigl(2+3x^{3}\bigr)^{3}} \ dx
\text{Put } t = 2 + 3x^{3} \implies dt = 9x^{2} \ dx \quad\Rightarrow\quad x^{2} \ dx = \frac{dt}{9}
I = \frac{1}{9}\int t^{-3} \ dt
I = \frac{1}{9}\cdot \left(\frac{t^{-2}}{-2}\right) + C
I = -\frac{1}{18}t^{-2} + C
I = -\frac{1}{18}\bigl(2+3x^{3}\bigr)^{-2} + C
Final Answer: \boxed{\int \frac{x^{2}}{\bigl(2+3x^{3}\bigr)^{3}} \ dx = -\frac{1}{18\bigl(2+3x^{3}\bigr)^{2}} + C}
Question 14: \int \frac{1}{x\bigl(\log x\bigr)^{m}} \ dx , \quad x>0 \ ; m\neq 1
Solution:
Let, I = \int \frac{1}{x\bigl(\log x\bigr)^{m}} \ dx
\text{Put } t = \log x \implies dt = \frac{dx}{x}
I = \int t^{-m} \ dt
I = \frac{t^{1-m}}{1-m} + C
I = \frac{(\log x)^{1-m}}{1-m} + C
Final Answer: \boxed{ \int \frac{1}{x\bigl(\log x\bigr)^{m}} \ dx = \frac{(\log x)^{1-m}}{1-m} + C}
Question 15: \int \frac{x}{9-4x^{2}} \ dx
Solution:
Let I = \int \frac{x}{9-4x^{2}} \ dx
\text{Put } t = 9 - 4x^{2} \implies dt = -8x \ dx \quad \Rightarrow \quad x \ dx = -\frac{dt}{8}
I = -\frac{1}{8}\int \frac{dt}{t}
I = -\frac{1}{8}\log|t| + C
I = -\frac{1}{8}\log(9-4x^{2}) + C
Final Answer: \boxed{\int \frac{x}{9-4x^{2}} \ dx = -\frac{1}{8}\log(9-4x^{2}) + C}
Question 16: \int e^{2x+3} \ dx
Solution:
Let, I = \int e^{2x+3} \ dx
\text{Put } t = 2x+3 \implies dt = 2 \ dx \quad \Rightarrow \quad dx = \frac{dt}{2}
I = \frac{1}{2}\int e^{t} \ dt
I = \frac{1}{2}e^{t} + C
I = \frac{1}{2}e^{2x+3} + C
Final Answer: \boxed{\int e^{2x+3} \ dx = \frac{1}{2}e^{2x+3} + C}
Question 17: \int \frac{x}{e^{x^{2}}} \ dx
Solution:
Let, I = \int \frac{x}{e^{x^{2}}} \ dx
I = \int x e^{-x^{2}} \ dx
\text{Put } t = x^{2} \implies dt = 2x \ dx \quad \Rightarrow \quad x \ dx = \frac{dt}{2}
I = \frac{1}{2}\int e^{-t} \ dt
I = -\frac{1}{2}e^{-t} + C
I = -\frac{1}{2}e^{-x^{2}} + C
Final Answer: \boxed{\int \frac{x}{e^{x^{2}}} \ dx = -\frac{1}{2e^{x^{2}}} + C}
Question 18: \int \frac{e^{\tan^{-1}x}}{1+x^{2}} \ dx
Solution:
Let, I = \int \frac{e^{\tan^{-1}x}}{1+x^{2}} \ dx
\text{Put } t = \tan^{-1}x \implies dt = \frac{dx}{1+x^{2}}
I = \int e^{t} dt
I= e^{t} + C
I = e^{\tan^{-1}x} + C
Final Answer: \boxed{\int \frac{e^{\tan^{-1}x}}{1+x^{2}} \ dx = e^{\tan^{-1}x} + C}
Question 19: \int \frac{e^{2x}-1}{e^{2x}+1} \ dx
Solution:
Let, I = \int \frac{e^{2x}-1}{e^{2x}+1} \ dx
dividing numerator and denominator by e^x
I = \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \ dx
\text{Put } t = e^{x}+e^{-x} \implies dt = (e^{x}-e^{-x}) \ dx
I = \int \frac {dt}{t}
I = \log|t| + C
I = \log|e^{x}+e^{-x}| + C
Final Answer: \boxed{\int \frac{e^{2x}-1}{e^{2x}+1} \ dx\ = log|e^{x}+e^{-x}| + C}
Question 20: \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} \ dx
Solution:
Let, I = \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} \ dx
\text{Put } t = e^{2x}+e^{-2x} \implies dt = (2e^{2x} - 2e^{-2x})dx \implies \frac{dt}{2} = (e^{2x} - e^{-2x}) dx
I = \frac{1}{2} \int \frac{dt}{t}
I = \frac{1}{2} \log{t} + C
I = \frac{1}{2} \log(e^{2x}+e^{-2x}) + C
Final Answer: \boxed{ \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} \ dx = \frac{1}{2} \log(e^{2x}+e^{-2x}) + C }
Question 21: \int \tan^{2}\bigl(2x-3\bigr) \ dx
Solution:
Let, I = \int \tan^{2}\bigl(2x-3\bigr) \ dx
\text{Use identity } \tan^{2}u=\sec^{2}u-1. \
I = \int (sec^{2}(2x-3) - 1) \ dx
\text{Put } u=2x-3, \quad du=2dx \Rightarrow dx=\frac{du}{2}.
I = \frac{1}{2}\int \bigl(\sec^{2}u-1\bigr) \ du
I = \frac{1}{2}\bigl(\tan u - u\bigr) + C
I = \frac{1}{2}\tan(2x-3) - \frac{1}{2}(2x-3) + C
I = \frac{1}{2}\tan(2x-3) - x + \frac{3}{2} + C
Final Answer: \boxed{\int \tan^{2}\bigl(2x-3\bigr) \ dx = \frac{1}{2}\tan(2x-3)-x+C}
Question 22: \int \sec^{2}\bigl(7-4x\bigr) \ dx
Solution:
Let, I = \int \sec^{2}\bigl(7-4x\bigr) \ dx
\text{Put } u=7-4x,\quad du=-4,dx\Rightarrow dx=-\frac{du}{4}.
I = -\frac{1}{4}\int \sec^{2}u \ du
I = -\frac{1}{4}\tan u + C
I= -\frac{1}{4}\tan(7-4x) + C
Final Answer: \boxed{\int \sec^{2}\bigl(7-4x\bigr) \ dx = -\tfrac{1}{4}\tan(7-4x)+C}
Question 23: \int \frac{\sin^{-1}x}{\sqrt{1-x^{2}}} \ dx \qquad(-1<x<1)
Solution:
Let, I = \int \frac{\sin^{-1}x}{\sqrt{1-x^{2}}} \ dx
\text{Put } u=\sin^{-1}x,\quad du=\frac{dx}{\sqrt{1-x^{2}}}
I = \int u \ du
I= \frac{u^{2}}{2}+C
I = \frac{\bigl(\sin^{-1}x\bigr)^{2}}{2}+C
Final Answer: \boxed{ \int \frac{\sin^{-1}x}{\sqrt{1-x^{2}}} \ dx = \frac{(\sin^{-1}x)^{2}}{2}+C}
Question 24: \int \frac{2\cos x - 3\sin x}{6\cos x + 4\sin x} \ dx
Let, I = \int \frac{2\cos x - 3\sin x}{6\cos x + 4\sin x} \ dx
\text{Put } t=6\cos x+4\sin x, \quad dt = (-6\sin x +4\cos x)dx.
\text{Observe } ;2\cos x-3\sin x=\tfrac{1}{2}(-6\sin x+4\cos x)=\tfrac{1}{2} \frac{dt}{dx}.
I = \frac{1}{2}\int \frac{dt}{t}
I = \frac{1}{2}\ln|t| + C
I = \frac{1}{2}\ln|6\cos x+4\sin x| + C
I = \frac{1}{2}[\ln|3\cos x+2\sin x| + \ln|2|] + C
\boxed{\int \frac{2\cos x - 3\sin x}{6\cos x + 4\sin x} \ dx = \frac{1}{2}\ln\bigl|3\cos x+2\sin x\bigr|+C}
Question 25: \int \frac{1}{\cos^{2}x\bigl(1-\tan x\bigr)^{2}} \ dx
Solution:
Let, I = \int \frac{1}{\cos^{2}x\bigl(1-\tan x\bigr)^{2}} \ dx
I = \int \frac{\sec^{2}x}{\bigl(1-\tan x\bigr)^{2}} \ dx
\text{Put } u=1-\tan x, \quad du=-\sec^{2}x \ dx
\text{Then } \sec^{2}x dx = -du
I = -\int \frac{du}{u^{2}}
I = -\bigl(-u^{-1}\bigr) + C
I = \frac{1}{u}+C
I = \frac{1}{1-\tan x} + C
Final Answer: \boxed{\int \frac{1}{\cos^{2}x\bigl(1-\tan x\bigr)^{2}} \ dx = \frac{1}{1-\tan x}+C}
Question 26: \int \frac{\cos\sqrt{x}}{\sqrt{x}} \ dx \quad x\ge 0
Solution:
Let, I = \int \frac{\cos\sqrt{x}}{\sqrt{x}} \ dx
\text{Put } u=\sqrt{x} \Rightarrow x=u^{2} \implies dx=2udu.
I = \int \frac{\cos u}{u}\cdot 2u,du
I = 2\int \cos u \ du
I = 2\sin u + C
I = 2\sin\bigl(\sqrt{x}\bigr) + C
Final Answer: \boxed{\int \frac{\cos\sqrt{x}}{\sqrt{x}} \ dx = 2\sin\bigl(\sqrt{x}\bigr)+C}
Question 27: \int \sqrt{\sin 2x} \cos 2x \ dx
Solution:
Let I = \int \sqrt{\sin 2x} \cos 2x \ dx
\text{Put } u=\sin 2x \Rightarrow du=2\cos 2x dx \Rightarrow \cos 2x dx = \tfrac{du}{2}
I = \frac{1}{2}\int u^{1/2} \ du
I = \frac{1}{2}\cdot \frac{2}{3}u^{3/2} + C
I = \frac{1}{3}u^{3/2} + C
I = \frac{1}{3}\bigl(\sin 2x\bigr)^{3/2} + C
Final Answer: \boxed{\int \sqrt{\sin 2x} \cos 2x \ dx = \tfrac{1}{3}\bigl(\sin 2x\bigr)^{3/2}+C}
Question 28: \int \frac{\cos x}{\sqrt{1+\sin x}} \ dx
Solution:
Let, I = \int \frac{\cos x}{\sqrt{1+\sin x}} \ dx
\text{Put } u=1+\sin x \implies du=\cos x \ dx
I = \int \frac{du}{\sqrt{u}}
I = 2\sqrt{u} + C
I = 2\sqrt{1+\sin x} + C
Final Answer: \boxed{\int \frac{\cos x}{\sqrt{1+\sin x}} \ dx = 2\sqrt{1+\sin x}+C}
Question 29: \int \cot x \ \log(\sin x) \ dx \qquad (0<x<\pi)
Solution:
Let, I = \int \cot x \ \log(\sin x) \ dx
\text{Put } u=\log(\sin x) \quad \Rightarrow \quad du=\cot x,dx.
I = \int u \ du
I = \frac{u^{2}}{2} + C
I = \frac{\bigl(\log(\sin x)\bigr)^{2}}{2} + C
Final Answer: \boxed{\int \cot x \ \log(\sin x) \ dx = \tfrac{1}{2}\bigl(\log(\sin x)\bigr)^{2}+C}
Question 30: \int \frac{\sin x}{1+\cos x} \ dx
Solution:
Let, I = \int \frac{\sin x}{1+\cos x} \ dx
\text{Put } u=1+\cos x \Rightarrow du=-\sin x dx \Rightarrow \sin x dx=-du
I = -\int \frac{du}{u}
I = -\ln|u| + C
I = -\ln(1+\cos x) + C
Final Answer: \boxed{\int \frac{\sin x}{1+\cos x} \ dx = -\ln(1+\cos x)+C}
Question 31: \int \frac{\sin x}{(1+\cos x)^{2}} \ dx
Solution:
Let, I= \int \frac{\sin x}{(1+\cos x)^{2}} \ dx
\text{Put } u=1+\cos x \Rightarrow du=-\sin x dx \Rightarrow \sin x dx=-du
I=-\int \frac{du}{u^{2}}
I=-\bigl(-u^{-1}\bigr)+C = u^{-1}+C
I=\frac{1}{1+\cos x}+C
Final Answer: \boxed{\int \frac{\sin x}{(1+\cos x)^{2}} \ dx = \frac{1}{1+\cos x}+C}
Question 32: \int \frac{1}{1+\cot x} \ dx
Solution:
Let, I= \int \frac{1}{1+\cot x} \ dx
\text{Rewrite } \frac{1}{1+\cot x}=\frac{1}{1+\frac{\cos x}{\sin x}}=\frac{\sin x}{\sin x+\cos x}
\text{So } I=\int \frac{\sin x}{\sin x+\cos x},dx
\text{Write } \frac{\sin x}{\sin x+\cos x}=1-\frac{\cos x}{\sin x+\cos x}
I=\int 1 \ dx - \int \frac{\cos x}{\sin x+\cos x} \ dx = x - \int \frac{\cos x}{\sin x+\cos x} \ dx
\text{Now } \frac{\cos x}{\sin x+\cos x}=\tfrac{1}{2}\left(\frac{\cos x-\sin x}{\sin x+\cos x}+\frac{\cos x+\sin x}{\sin x+\cos x}\right)
\Rightarrow \int \frac{\cos x}{\sin x+\cos x} \ dx = \tfrac{1}{2}\int \frac{\cos x-\sin x}{\sin x+\cos x} \ dx + \tfrac{1}{2}\int dx
\text{Put } u=\sin x+\cos x \Rightarrow du=(\cos x-\sin x) \ dx
\Rightarrow \tfrac{1}{2}\int \frac{\cos x-\sin x}{\sin x+\cos x} \ dx = \tfrac{1}{2}\int \frac{du}{u} = \tfrac{1}{2}\ln|u| + C
\therefore \int \frac{\cos x}{\sin x+\cos x} \ dx = \tfrac{1}{2}\ln|\sin x+\cos x| + \tfrac{1}{2}x + C
\text{Hence } I = x - \Bigl(\tfrac{1}{2}\ln|\sin x+\cos x| + \tfrac{1}{2}x\Bigr) + C
I=\tfrac{1}{2}x - \tfrac{1}{2}\ln|\sin x+\cos x| + C
Final Answer: \boxed{\displaystyle \int \frac{1}{1+\cot x} \ dx=\frac{x}{2}-\frac{1}{2}\ln\bigl|\sin x+\cos x\bigr|+C}
Question 33: \int \frac{1}{1-\tan x} \ dx
Solution:
Let, I= \int \frac{1}{1-\tan x} \ dx
\text{Rewrite } \frac{1}{1-\tan x}=\frac{1}{1-\frac{\sin x}{\cos x}}=\frac{\cos x}{\cos x-\sin x}
\text{So } I=\int \frac{\cos x}{\cos x-\sin x},dx
\text{Write } \frac{\cos x}{\cos x-\sin x}=\tfrac{1}{2}\left(\frac{\cos x+\sin x}{\cos x-\sin x}+\frac{\cos x-\sin x}{\cos x-\sin x}\right)
\Rightarrow \frac{\cos x}{\cos x-\sin x}=\tfrac{1}{2}\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)+\tfrac{1}{2}
I=\frac{1}{2}\int \frac{\cos x+\sin x}{\cos x-\sin x},dx + \tfrac{1}{2}\int dx
\text{Put } u=\cos x-\sin x \Rightarrow du = -( \sin x +\cos x)dx = -(\cos x+\sin x)dx
\Rightarrow \int \frac{\cos x+\sin x}{\cos x-\sin x},dx = -\int \frac{du}{u} = -\ln|u| + C
\therefore I = \tfrac{1}{2}\bigl(-\ln|\cos x-\sin x|\bigr) + \tfrac{1}{2}x + C
I=\tfrac{x}{2} - \tfrac{1}{2}\ln|\cos x-\sin x| + C
Final Answer: \boxed{ \int \frac{1}{1-\tan x} \ dx=\frac{x}{2}-\frac{1}{2}\ln\bigl|\cos x-\sin x\bigr|+C}
Question 34: \int \frac{\sqrt{\tan x}}{\sin x\cos x} \ dx
Solution:
Let, I= \int \frac{\sqrt{\tan x}}{\sin x\cos x} \ dx
Dividing numerator and denominator by \cos^{2} x
I= \int \frac{\sec^{2}x \sqrt{\tan x}}{\tan x} \ dx
I= \int \frac{\sec^{2}x} {\sqrt{\tan x}} \ dx
\text{Put } u=\tan x \Rightarrow du=\sec^{2}x \ dx
I = \int \frac{du}{\sqrt{u}}
I=\int u^{-1/2} \ du
I = 2u^{1/2} + C
I=2\sqrt{\tan x}+C
Final Answer: \boxed{\displaystyle \int \frac{\sqrt{\tan x}}{\sin x\cos x},dx=2\sqrt{\tan x}+C}
Question 35: \int \frac{\bigl(1+\log x\bigr)^{2}}{x},dx,\quad x>0
Solution:
Let, I=\displaystyle \int \frac{(1+\log x)^{2}}{x},dx
\text{Put } u=1+\log x \Rightarrow du=\frac{dx}{x}
I=\int u^{2}\ du
I= \frac{u^{3}}{3}+C
I=\frac{(1+\log x)^{3}}{3}+C
Final Answer: \boxed{\int \frac{\bigl(1+\log x\bigr)^{2}}{x} \ dx=\frac{(1+\log x)^{3}}{3}+C}
Question 36: \int \frac{(x+1)\bigl(x+\log x\bigr)^{2}}{x},dx,\quad x>0
Solution:
Let, I= \int \frac{(x+1)\bigl(x+\log x\bigr)^{2}}{x} \ dx
\text{Write } \frac{x+1}{x}=1+\frac{1}{x}
I=\int \Bigl(1+\frac{1}{x}\Bigr)\bigl(x+\log x\bigr)^{2} \ dx
\text{Put } u = x+\log x \Rightarrow du=\Bigl(1+\frac{1}{x}\Bigr) \ dx
I=\int u^{2} \ du
I = \frac{u^{3}}{3}+C
I=\frac{\bigl(x+\log x\bigr)^{3}}{3}+C
Final Answer: \boxed{\int \frac{(x+1)\bigl(x+\log x\bigr)^{2}}{x} \ dx=\frac{\bigl(x+\log x\bigr)^{3}}{3}+C}
Question 37: \int \frac{x^{3}\sin\bigl(\tan^{-1}x^{4}\bigr)}{1+x^{8}} \ dx
Solution:
Let, I=\displaystyle \int \frac{x^{3}\sin\bigl(\tan^{-1}x^{4}\bigr)}{1+x^{8}},dx
\text{Put } u=\tan^{-1}x^{4} \Rightarrow du=\frac{4x^{3}}{1+x^{8}}dx \Rightarrow \frac{x^{3}}{1+x^{8}}dx=\frac{1}{4}du
I=\frac{1}{4}\int \sin u \ du = -\frac{1}{4}\cos u + C
I=-\frac{1}{4}\cos\bigl(\tan^{-1}x^{4}\bigr)+C
Final Answer: \boxed{\int \frac{x^{3}\sin\bigl(\tan^{-1}x^{4}\bigr)}{1+x^{8}},dx=-\frac{1}{4}\cos\bigl(\tan^{-1}x^{4}\bigr)+C}
Question 38: \int \frac{10x^{9}+10^{x}\ln 10}{x^{10}+10^{x}} \ dx \qquad\text{equals}
Solution:
Let, I= \int \frac{10x^{9}+10^{x}\ln 10}{x^{10}+10^{x}},dx
\text{Put } u=x^{10}+10^{x} \Rightarrow du = 10x^{9} \ dx + 10^{x}\ln 10 \ dx
I = \int \frac{du}{u}
I = \ln|u| + C
I=\ln\bigl(x^{10}+10^{x}\bigr)+C
\boxed{\int \frac{10x^{9}+10^{x}\ln 10}{x^{10}+10^{x}},dx=\ln\bigl(x^{10}+10^{x}\bigr)+C}
Correct Option: (D)
Question 39: \int \frac{dx}{\sin^{2}x\cos^{2}x} \ dx \qquad\text{equals}
Solution:
Let, I= \int \frac{dx}{\sin^{2}x\cos^{2}x} \ dx
\text{Note } \frac{1}{\sin^{2}x\cos^{2}x}=\frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos^{2}x} =\csc^{2}x+\sec^{2}x
I=\int \bigl(\csc^{2}x+\sec^{2}x\bigr) \ dx
I = -\cot x + \tan x + C
\boxed{ \int \frac{dx}{\sin^{2}x\cos^{2}x} \ dx=\tan x-\cot x + C}
Correct Option: (B)