Class 12 NCERT Maths – Chapter 7, Exercise 7.1 Solutions

Find an anti-derivative (or integral) of the following functions (Questions 1 to 5) by the method of inspection

Question 1: Find \int \sin 2x \ dx

Solution:
I = \int \sin 2x \ dx

We know,
\frac{d}{dx}(\cos 2x) = -2 \sin 2x

So,
\sin 2x = -\frac{1}{2} \cdot \frac{d}{dx}(\cos 2x)

Therefore,
I = -\frac{1}{2} \cos 2x + C

Final Answer: \boxed{\int \sin 2x \ dx = -\frac{1}{2} \cos 2x + C }

Question 2: Find \int \cos 3x \ dx

Solution:
I = \int \cos 3x \ dx

We know,
\frac{d}{dx}(\sin 3x) = 3 \cos 3x

So,
\cos 3x = \frac{1}{3} \cdot \frac{d}{dx}(\sin 3x)

Therefore,
I = \frac{1}{3} \sin 3x + C

Final Answer: \boxed{ \int \cos 3x , dx = \frac{1}{3} \sin 3x + C }

Question 3: Find \int e^{2x} \ dx

Solution:
I = \int e^{2x} \ dx

We know,
\frac{d}{dx}(e^{2x}) = 2e^{2x}

So,
e^{2x} = \tfrac{1}{2} \cdot \frac{d}{dx}(e^{2x})

Therefore,
I = \tfrac{1}{2} e^{2x} + C

Final Answer: \boxed{ \int e^{2x} \ dx = \tfrac{1}{2} e^{2x} + C }

Question 4: Find \int (ax + b)^2 \ dx

Solution:
I = \int (ax + b)^2 \ dx

Let t = ax + b \Rightarrow dt = a , dx \Rightarrow dx = \tfrac{1}{a} dt

Thus,
I = \int (t^2) \cdot \tfrac{1}{a} dt

I = \tfrac{1}{a} \int t^2 dt

I = \tfrac{1}{a} \cdot \tfrac{t^3}{3} + C

I = \tfrac{(ax+b)^3}{3a} + C

Final Answer: \boxed{ \int (ax + b)^2 \ dx = \tfrac{(ax+b)^3}{3a} + C }

Question 5: Find \int (\sin 2x - 4 e^{3x}) \ dx

Solution:
I = \int (\sin 2x - 4 e^{3x}) \ dx

Split the integral:
I = \int \sin 2x \ dx - 4 \int e^{3x} \ dx

From Q1,
\int \sin 2x \ dx = -\tfrac{1}{2} \cos 2x

Also,
\int e^{3x} \ dx = \tfrac{1}{3} e^{3x}

Therefore,
I = -\tfrac{1}{2} \cos 2x - 4 \cdot \tfrac{1}{3} e^{3x} + C

I = -\tfrac{1}{2} \cos 2x - \tfrac{4}{3} e^{3x} + C

Final Answer: \boxed{ \int (\sin 2x - 4 e^{3x}) \ dx = -\tfrac{1}{2} \cos 2x - \tfrac{4}{3} e^{3x} + C }

Find the following integrals in Question 6 to 20

Question 6: Evaluate \int\bigl(4e^{3x}+1\bigr) \ dx

Solution:
Let, I=\int\bigl(4e^{3x}+1\bigr) \ dx

I =4\int e^{3x},dx+\int 1 \ dx

I =4\cdot\frac{1}{3}e^{3x}+x+C

Final Answer: \boxed{ \int\bigl(4e^{3x}+1\bigr) = \frac{4}{3}e^{3x}+x+C}

Question 7: Evaluate \int x^{2}\Bigl(1-\frac{1}{x^{2}}\Bigr) \ dx

Solution:
Let, I = \int x^{2}\Bigl(1-\frac{1}{x^{2}}\Bigr) \ dx

I =\int\bigl(x^{2}-1\bigr) \ dx

I =\int x^{2} \ dx - \int 1 \ dx

I =\frac{x^{3}}{3}-x + C

Final Answer: \boxed{\int x^{2}\Bigl(1-\frac{1}{x^{2}}\Bigr) \ dx = \dfrac{x^{3}}{3}-x+C}

Question 8: Evaluate \int\bigl( ax^{2}+bx+c \bigr) \ dx \quad (\text{where }a,b,c\text{ are constants})

Solution:
Let, I = \int\bigl(ax^{2}+bx+c\bigr) \ dx

I =\int ax^{2} \ dx+\int bx \ dx+\int c \ dx

I = a\cdot\frac{x^{3}}{3} + b\cdot\frac{x^{2}}{2} + cx+ C

Final Answer: \boxed{\int\bigl(ax^{2}+bx+c\bigr) \ dx = \dfrac{a x^{3}}{3}+\dfrac{b x^{2}}{2}+c x+C}

Question 9: Evaluate \int\bigl(2x^{2}+e^{x}\bigr) \ dx

Solution:
I = \int\bigl(2x^{2}+e^{x}\bigr) \ dx

I = \int 2x^{2},dx+\int e^{x} \ dx

I = 2\cdot\frac{x^{3}}{3}+e^{x}+C

Final Answer: I = \boxed{\dfrac{2}{3}x^{3}+e^{x}+C}

Question 10: Evaluate \int\Bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)^{2} \ dx

Solution:
Let, I = \int\Bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)^{2} \ dx

\text{First expand:} \quad \Bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)^{2} = x - 2 + \frac{1}{x}

I =\int\Bigl(x-2+\frac{1}{x}\Bigr) \ dx

I =\frac{x^{2}}{2}-2x+\ln\lvert x\rvert +C

Final Answer: \boxed{ \int\Bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)^{2} \ dx = \dfrac{x^{2}}{2}-2x+\ln\lvert x\rvert +C}

Question 11: Evaluate \int\frac{x^{3}+5x^{2}-4}{x^{2}} \ dx

Solution:
Let, I = \int\frac{x^{3}+5x^{2}-4}{x^{2}} \ dx

\text{Divide termwise:} \quad \frac{x^{3}+5x^{2}-4}{x^{2}} =x+5-4x^{-2}

I =\int\bigl(x+5-4x^{-2}\bigr) \ dx

I =\frac{x^{2}}{2}+5x+4x^{-1}+C

Final Answer: \boxed{\int\frac{x^{3}+5x^{2}-4}{x^{2}} \ dx = \frac{x^{2}}{2}+5x+\dfrac{4}{x}+C}

Question 12: Evaluate \int\frac{x^{3}+3x+4}{\sqrt{x}} \ dx

Solution:
I = \int\frac{x^{3}+3x+4}{\sqrt{x}} \ dx

\text{Write powers: } \quad \frac{x^{3}+3x+4}{\sqrt{x}} = x^{\tfrac{5}{2}}+3x^{\tfrac{1}{2}}+4x^{-\tfrac{1}{2}}

I =\int\bigl(x^{\tfrac{5}{2}}+3x^{\tfrac{1}{2}}+4x^{-\tfrac{1}{2}}\bigr) \ dx

I =\frac{2}{7}x^{\tfrac{7}{2}}+2x^{\tfrac{3}{2}}+8x^{\tfrac{1}{2}}+C

Final Answer: \boxed{ \int\frac{x^{3}+3x+4}{\sqrt{x}} \ dx = \frac{2}{7}x^{\tfrac{7}{2}}+2x^{\tfrac{3}{2}}+8x^{\tfrac{1}{2}}+C}

Question 13: Evaluate \int\frac{x^{3}-x^{2}+x-1}{x-1} \ dx

Solution:
Let, I = \int\frac{x^{3}-x^{2}+x-1}{x-1} \ dx

\text{Factor numerator by grouping:}\quad

x^{3}-x^{2}+x-1 = (x^{3}-x^{2})+(x-1)

x^{2}(x-1)+1\cdot(x-1)=(x-1)(x^{2}+1)

\therefore \quad \frac{x^{3}-x^{2}+x-1}{x-1} = \frac{(x-1)(x^{2}+1)}{(x - 1)} = x^{2}+1

I =\int\bigl(x^{2}+1\bigr) \ dx

I =\frac{x^{3}}{3}+x+C

Final Answer: \boxed{\int\frac{x^{3}-x^{2}+x-1}{x-1} \ dx = \dfrac{x^{3}}{3}+x+C}

Question 14: Evaluate \int(1-x)\sqrt{x} \ dx

Solution:
Let, I = \int(1-x)\sqrt{x} \ dx

(1-x)\sqrt{x}=x^{\tfrac{1}{2}}-x^{\tfrac{3}{2}}

I = \int\bigl(x^{\tfrac{1}{2}}-x^{\tfrac{3}{2}}\bigr) \ dx

I =\frac{2}{3}x^{\tfrac{3}{2}}-\frac{2}{5}x^{\tfrac{5}{2}}+C

Final Answer: \boxed{\int(1-x)\sqrt{x} \ dx = \dfrac{2}{3}x^{\tfrac{3}{2}}-\dfrac{2}{5}x^{\tfrac{5}{2}}+C}

Question 15: Evaluate \int\sqrt{x} \bigl(3x^{2}+2x+3\bigr) \ dx

Solution:
Let, I = \int\sqrt{x} \bigl(3x^{2}+2x+3\bigr) \ dx

\text{Expand with powers:} \quad \sqrt{x}(3x^{2}+2x+3)=3x^{\tfrac{5}{2}}+2x^{\tfrac{3}{2}}+3x^{\tfrac{1}{2}}

I =\int\bigl(3x^{\tfrac{5}{2}}+2x^{\tfrac{3}{2}}+3x^{\tfrac{1}{2}}\bigr) \ dx

I =3\cdot\frac{2}{7}x^{\tfrac{7}{2}}+2\cdot\frac{2}{5}x^{\tfrac{5}{2}}+3\cdot\frac{2}{3}x^{\tfrac{3}{2}}+C

I =\frac{6}{7}x^{\tfrac{7}{2}}+\frac{4}{5}x^{\tfrac{5}{2}}+2x^{\tfrac{3}{2}}+C

Final Answer: \boxed{\int\sqrt{x} \bigl(3x^{2}+2x+3\bigr) \ dx = \dfrac{6}{7}x^{\tfrac{7}{2}}+\dfrac{4}{5}x^{\tfrac{5}{2}}+2x^{\tfrac{3}{2}}+C}

Question 16: Evaluate \int(2x-3\cos x+e^{x}) \ dx

Solution:
Let, I=\int(2x-3\cos x+e^{x}) \ dx

I=\int 2x \ dx - \int 3\cos x \ dx + \int e^{x} \ dx

I=x^{2}-3\sin x+e^{x}+C

Final Answer: \boxed{ \int(2x-3\cos x+e^{x}) \ dx = x^{2}-3\sin x+e^{x}+C}

Question 17: Evaluate \int (2x^{2}-3\sin x+5\sqrt{x}) \ dx

Solution:
Let, I=\int (2x^{2}-3\sin x+5\sqrt{x}) \ dx

I= \int 2x^{2} \ dx-\int 3\sin x \ dx+\int 5x^{\tfrac{1}{2}} \ dx

I= \frac{2}{3}x^{3}+3\cos x+\frac{10}{3}x^{\tfrac{3}{2}}+C

Final Answer: \boxed{ \int (2x^{2}-3\sin x+5\sqrt{x}) \ dx = \frac{2}{3}x^{3}+3\cos x+\frac{10}{3}x^{\frac{3}{2}}+C}

Question 18: Evaluate \int \sec x(\sec x+\tan x) dx

Solution:
Let, I=\int \sec x(\sec x + \tan x) \ dx

I=\int \sec^{2}x \ dx + \int \sec x\tan x \ dx

I=\tan x+\sec x+C

Final Answer: \boxed{\int \sec x(\sec x + \tan x) \ dx = \tan x+\sec x+C}

Question 19: Evaluate: \int \dfrac{\sec^{2}x}{\csc^{2}x} \ dx

Solution:
Let, I=\int \dfrac{\sec^{2}x}{\csc^{2}x} \ dx

\dfrac{\sec^{2}x}{\csc^{2}x}=\dfrac{1/\cos^{2}x}{1/\sin^{2}x}=\dfrac{\sin^{2}x}{\cos^{2}x}=\tan^{2}x

I=\int \tan^{2}x \ dx

\tan^{2}x=\sec^{2}x-1

I=\int(\sec^{2}x-1) \ dx

I =\tan x-x+C

Final Answer: \boxed{\int \dfrac{\sec^{2}x}{\csc^{2}x} \ dx = \tan x-x+C}

Question 20: Evaluate: \int \dfrac{2-3\sin x}{\cos^{2}x} \ dx

Solution:
Let, I=\int \dfrac{2-3\sin x}{\cos^{2}x} \ dx

I=\int \dfrac{2}{\cos^{2}x} \ dx-\int \dfrac{3\sin x}{\cos^{2}x} \ dx

I=2\int \sec^{2}x \ dx-3\int \dfrac{\sin x}{\cos^{2}x} \ dx

I=2\tan x-3\int \tan x\sec x \ dx

I=2\tan x-3\sec x+C

Final Answer: \boxed{\int \dfrac{2-3\sin x}{\cos^{2}x} \ dx = 2\tan x-3\sec x+C}

Question 21: The anti-derivative of \int\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right) \ dx equals

(A) \tfrac{1}{3}x^{\tfrac{3}{2}}+2x^{\tfrac{1}{2}}+C

(B) \tfrac{2}{3}x^{\tfrac{3}{2}}+\tfrac{1}{2}x^{2}+C

(C) \tfrac{2}{3}x^{\tfrac{3}{2}}+2x^{\tfrac{1}{2}}+C

(D) \tfrac{3}{2}x^{\tfrac{3}{2}}+\tfrac{1}{2}x^{2}+C

Solution:
I=\int\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right) \ dx

I=\int x^{\tfrac{1}{2}} \ dx+\int x^{-\tfrac{1}{2}} \ dx

I=\dfrac{2}{3}x^{\tfrac{3}{2}}+2x^{\tfrac{1}{2}}+C

\boxed{\int\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right) \ dx = \dfrac{2}{3}x^{\tfrac{3}{2}}+2x^{\tfrac{1}{2}}+C}

Correct Option: (C)

Question 22: If \dfrac{d}{dx}f(x)=4x^{3}-\dfrac{3}{x^{4}} such that f(2) = 0 then f(x) is

(A) x^{4}+\dfrac{1}{x^{3}}-\dfrac{129}{8}

(B) x^{3}+\dfrac{1}{x^{4}}+\dfrac{129}{8}

(C) x^{4}+\dfrac{1}{x^{3}}+\dfrac{129}{8}

(D) x^{3}+\dfrac{1}{x^{4}}-\dfrac{129}{8}

Solution:
f(x)=\int\left(4x^{3}-\dfrac{3}{x^{4}}\right) \ dx

f(x)=\int 4x^{3} \ dx-\int 3x^{-4} \ dx

f(x)=x^{4}+x^{-3}+C

f(x)=x^{4}+\dfrac{1}{x^{3}}+C

f(2)=0\Rightarrow 0=16+\dfrac{1}{8}+C

C=-\dfrac{129}{8}

f(x)=x^{4}+\dfrac{1}{x^{3}}-\dfrac{129}{8}

\boxed{f(x) = x^{4}+\dfrac{1}{x^{3}}-\dfrac{129}{8}}

Correct Option: (A)