Class 12 Integration Exercise 7.8 NCERT Solutions

This exercise is based on formulas for integration, before you start the exercise it is advised that you go through the formulas from the following link. Important integration formulas

Question 1: Evaluate: \int_{-1}^{1} (x + 1) \ dx

Solution:
\begin{aligned} I = \int_{-1}^{1} (x + 1) \ dx \end{aligned}

I = \Big[ \tfrac{x^{2}}{2} + x \Big]_{-1}^{1}

I = \Big( \tfrac{1}{2} + 1 \Big) - \Big( \tfrac{1}{2} - 1 \Big)

I = 2.\

Final Answer: \boxed{ \int_{-1}^{1} (x + 1) \ dx = 2}

Question 2: Evaluate: \int_{2}^{3} \dfrac{1}{x} \ dx

Solution:
I = \int_{2}^{3} \dfrac{1}{x} dx

I = [ \ln x ]_{2}^{3}

I = ln 3 - ln 2

I = \ln \dfrac{3}{2}. \

Final Answer: \boxed{ \int_{2}^{3} \dfrac{1}{x} dx = \ln \dfrac{3}{2}}

Question 3: \int_{1}^{2} \big(4x^{3} - 5x^{2} + 6x + 9\big) \ dx

Solution:
I = \int_{1}^{2} (4x^{3} - 5x^{2} + 6x + 9) \ dx

I = \Big[ x^{4} - \tfrac{5}{3}x^{3} + 3x^{2} + 9x \Big]_{1}^{2}

I = [ 2^{4} - \tfrac{5}{3}\cdot2^{3} + 3\cdot2^{2} + 9\cdot2 \Big] - [ 1^{4} - \tfrac{5}{3}\cdot1^{3} + 3\cdot1^{2} + 9\cdot1 \Big]

I = \dfrac{64}{3}

Final Answer: \boxed{\int_{1}^{2} (4x^{3} - 5x^{2} + 6x + 9) \ dx = \dfrac{64}{3}}

Question 4: \int_{0}^{\tfrac{\pi}{4}} \sin 2x \ dx

Solution:
I = \int_{0}^{\tfrac{\pi}{4}} \sin 2x \ dx

I = \Big[ -\tfrac{1}{2}\cos 2x \Big]_{0}^{\pi/4}

I = 0 - \Big( -\tfrac{1}{2} \Big)

I = \tfrac{1}{2}

Final Answer: \boxed{\int_{0}^{\tfrac{\pi}{4}} \sin 2x \ dx = \tfrac{1}{2}}

Question 5: \int_{0}^{\tfrac{\pi}{2}} \cos 2x \ dx

Solution:
I = \int_{0}^{\tfrac{\pi}{2}} \cos 2x \ dx

I = \Big[ \tfrac{1}{2}\sin 2x \Big]_{0}^{\pi/2}

I = \Big[ \tfrac{1}{2}\sin (2\cdot \frac{\pi}{2}) \Big] - \Big[ \tfrac{1}{2}\sin 2\cdot 0 \Big]

I = \Big[ \tfrac{1}{2}\sin (\pi) \Big] - \Big[ \tfrac{1}{2}\sin 0 \Big]

I = 0 - 0

Final Answer: \boxed{ I = 0}

Question 6: \int_{4}^{5} e^{x} \ dx

Solution:
I = \int_{4}^{5} e^{x} dx

I = [ e^{x} ]_{4}^{5}

I = e^{5} - e^{4}

Final Answer: \boxed{I = e^{4} (e - 1)}

Question 7: \int_{0}^{\tfrac{\pi}{4}} \tan x , dx

Solution:
I = \int_{0}^{\tfrac{\pi}{4}} \tan x \ dx

I = [ \ln \sec x ]_{0}^{\pi/4}

I = \ln |\sec(\pi/4)| - \ln |\sec(0)|

I= \ln \sqrt{2} - \ln 1

I = \tfrac{1}{2}\ln 2 - 0

I = \tfrac{1}{2}\ln 2

Final Answer: \boxed{\int_{0}^{\tfrac{\pi}{4}} \tan x \ dx = \tfrac{1}{2}\ln 2}

Question 8: \int_{\tfrac{\pi}{6}}^{\tfrac{\pi}{4}} \cosec x \ dx

Solution:
I = \int_{\tfrac{\pi}{6}}^{\tfrac{\pi}{4}} \cosec x \ dx

I = \Big [log |csc(x) - cot (x)| \Big ]_{\frac{\pi}{6}}^{\frac{\pi}{4}}

I = \Big [log |csc(\frac{\pi}{4}) - cot (\frac{\pi}{4})| \Big ] - \Big [log |csc(\frac{\pi}{6}) - cot (\frac{\pi}{6})| \Big ]

I = log |\sqrt {2} - 1| - log |2 - \sqrt {3}|

I = log \Big | \tfrac{\sqrt {2} - 1}{2 - \sqrt {3}} \Big |

Final Answer: \boxed{ \int_{\tfrac{\pi}{6}}^{\tfrac{\pi}{4}} \cosec x \ dx = log \Big | \tfrac{\sqrt {2} - 1}{2 - \sqrt {3}} \Big | }

Question 9: \int_{0}^{1} \dfrac{dx}{\sqrt{1 - x^{2}}}​

Solution:
I = \int_{0}^{1} \dfrac{dx}{\sqrt{1 - x^{2}}}

I = [ \sin^{-1} x ]_{0}^{1}

I = [ \sin^{-1} 1] - [ \sin^{-1} 0]

I = \tfrac{\pi}{2} - 0

Final Answer: \boxed{ \int_{0}^{1} \dfrac{dx}{\sqrt{1 - x^{2}}} = \tfrac{\pi}{2} }

Question 10: \int_{0}^{1} \dfrac{dx}{1 + x^{2}}

Solution:
I = \int_{0}^{1} \dfrac{dx}{1 + x^{2}}

I = [ \tan^{-1} x ]_{0}^{1}

I = [ \tan^{-1} 1 ] - [ \tan^{-1} 0 ]

I = \tfrac{\pi}{4} - 0

Final Answer: \boxed{ \int_{0}^{1} \dfrac{dx}{1 + x^{2}} = \tfrac{\pi}{4}}

Question 11: \int_{2}^{3} \dfrac{dx}{x^{2} - 1}​

Solution:
I = \int_{2}^{3} \dfrac{dx}{x^2-1} \ dx
​
I= \int_{2}^{3} \dfrac{dx}{(x - 1)(x + 1)} \ dx

I = \int_{2}^{3} \dfrac{1}{2}\Big( \dfrac{1}{x - 1} - \dfrac{1}{x + 1} \Big) \ dx

I = \dfrac{1}{2} \int_{2}^{3} \Big( \dfrac{1}{x - 1} - \dfrac{1}{x + 1} \Big) \ dx

I = \dfrac{1}{2} \Big[ log(x - 1) - log(x + 1) \Big]_{2}^{3}

I = \dfrac{1}{2} \Big[ (log(2) - log(4)) - ( log (1) - log (3)) \Big]

I = \dfrac{1}{2} \Big[ (log(2) - 2 \cdot log(2)) - ( - log (3)) \Big]

I = \dfrac{1}{2} \Big[ (log(3) - log(2)) \Big]

I = \dfrac{1}{2} \cdot log \Big( \frac{3}{2} \Big)

Final Answer: \boxed{ \int_{2}^{3} \dfrac{dx}{x^2-1} \ dx = \dfrac{1}{2} \cdot log \frac{3}{2} }

Question 12: \int_{0}^{\tfrac{\pi}{2}} \cos^{2} x \ dx

Solution:
I = \int_{0}^{\tfrac{\pi}{2}} \cos^{2} x \ dx

I = \int_{0}^{\pi/2} \tfrac{1}{2}(1+\cos 2x) \ dx

I = \Big[ \tfrac{x}{2} + \tfrac{\sin 2x}{4} \Big]_{0}^{\pi/2}

I = \Big[ \tfrac{\pi}{4} + \tfrac{\sin \pi}{4} \Big] - \Big[ \tfrac{0}{4} + \tfrac{\sin 0}{4} \Big]

I = \tfrac{\pi}{4}

Final Answer: \boxed{ \int_{0}^{\tfrac{\pi}{2}} \cos^{2} x \ dx = \dfrac{\pi}{4}}

Question 13: \int_{2}^{3} \dfrac{x}{x^{2}+1} \ dx

Solution:
I = \int_{2}^{3} \dfrac{x}{x^{2}+1} \ dx

Let x^2 + 1 = t \implies 2x = \dfrac{dt}{dx} \implies x \cdot dx = \dfrac {dt}{2}

\int \dfrac{x}{x^{2}+1} \ dx = \dfrac {1}{2} \int \dfrac{dt}{t}

\int \dfrac{x}{x^{2}+1} \ dx = \dfrac {1}{2} \log t

Substituting back t

\int \dfrac{x}{x^{2}+1} \ dx = \dfrac {1}{2} \log |x^2 + 1|

I = \Big [ \dfrac {1}{2} \log |x^2 + 1| \Big]_{2}^{3}

I = \dfrac {1}{2} \Big [ (\log 10 - \log 5) \Big]

Final Answer: \boxed { I = \dfrac {1}{2} \Big ( \log 2 \Big) }

Question 14: \int_{0}^{1} \dfrac{2x+3}{5x^{2}+1}, dx

I = \int_{0}^{1} \dfrac{2x+3}{5x^{2}+1} \ dx

Let, 2x + 3 = A\cdot \dfrac {d(5x^2 + 1)}{dx} + B

2x + 3 = A\cdot 10x + B

Comparing coefficients: A = \frac {1}{5} \ \ B = 3

2x + 3 = \dfrac{1}{5}\cdot 10x + 3

I = \int_{0}^{1} \dfrac{\dfrac{1}{5}\cdot 10x + 3}{5x^{2}+1} \ dx

I = \int_{0}^{1} \dfrac{\dfrac{1}{5}\cdot 10x + 3}{5x^{2}+1} \ dx

I = \int_{0}^{1} \Big[ \dfrac{10x}{5 \cdot (5x^{2}+1)} + \dfrac{3}{5x^{2}+1} \Big] \ dx

I = \int_{0}^{1} \dfrac{10x}{5 \cdot (5x^{2}+1)} \ dx + \int_{0}^{1} \dfrac{3}{5x^{2}+1} \ dx

Let, I_1 = \int \dfrac{10x}{5 \cdot (5x^{2}+1)} \ dx

Let, 5x^2 + 1 = t \implies 10x = \dfrac{dt}{dx} \implies 10x \cdot dx = dt

I_1 = \dfrac{1}{5} \int \dfrac{dt}{t} \ dx

I_1 = \dfrac{1}{5} \cdot log t

Substituting back t,

I_1 = \dfrac{1}{5} \cdot log |5x^2+1|

Let, I_2 = \int \dfrac{3}{5x^{2}+1} \ dx

I_2 = 3 \cdot \int \dfrac{1}{(x \sqrt{5})^2 + 1^2} \ dx

I_2 = \dfrac {3}{\sqrt{5}} \cdot \tan^{-1} (x\sqrt{5})

Therefore I = \Big [\dfrac{1}{5} \cdot log |5x^2+1| \Big]_{0}^{1} + \Big [\dfrac {3}{\sqrt{5}} \cdot \tan^{-1} (x\sqrt{5}) \Big]_{0}^{1}

I = \dfrac{1}{5} \big[ log 6 - log 1 ] \Big] + \dfrac {3}{\sqrt{5}} \Big [ \tan^{-1} \sqrt{5} - \tan^{-1} 0 \Big]

I = \dfrac{1}{5} log 6 + \dfrac {3}{\sqrt{5}} \tan^{-1} \sqrt{5}

Final Answer: \boxed {\int_{0}^{1} \dfrac{2x+3}{5x^{2}+1} \ dx = \dfrac{1}{5} log 6 + \dfrac {3}{\sqrt{5}} \tan^{-1} \sqrt{5}}

Question 15: \int_{0}^{1} x e^{x^{2}} \ dx

Solutions:
Let, I = \int_{0}^{1} x e^{x^{2}} \ dx

Let, t = x^{2} \implies dt = 2xdx

\int x e^{x^{2}} \ dx = \tfrac{1}{2}\int e^{t} \ dt

\int x e^{x^{2}} \ dx = \tfrac{1}{2} e^{t}

Substituting back x
\int x e^{x^{2}} \ dx = \tfrac{1}{2} e^{x}

I = \tfrac{1}{2} \Big [ e^{x}]_{0}^{1}

I = \tfrac{1}{2} \Big [ e^{1} - e{0}]

I = \tfrac{1}{2} (e - 1)

Final Answer: \boxed{ \int_{0}^{1} x e^{x^{2}} \ dx = \frac{1}{2}(e-1)}

Question 16: \int_{1}^{2} \dfrac{5x^{2}}{x^{2}+4x+3} \ dx

Solutions:
Let, I = \int_{1}^{2} \dfrac{5x^{2}}{x^{2}+4x+3} \ dx

\dfrac{5x^{2}}{x^{2}+4x+3} = 5 - \dfrac{20x+15}{x^{2}+4x+3}

I = \int_{1}^{2} 5 - \frac{20x+15}{x^{2}+4x+3} \ dx

I = \int_{1}^{2} 5 \ dx - \int_{1}^{2} \frac{20x+15}{x^{2}+4x+3} \ dx

I = I_1 - I_2

Let, I_1 = \int_{1}^{2} 5 \ dx

I_1 = \Big[ 5x \Big]_{1}^{2} \ dx

I_1 = (10) - (5) = 5

I_2 = \int_{1}^{2} \frac{20x+15}{x^{2}+4x+3} \ dx

\frac{20x+15}{x^2+4x+3} = \frac{20x + 15}{(x+1)(x+3)}

\frac{20x+15}{x^2+4x+3} = \frac{A}{(x+1)} + \frac{B}{(x+3)}

\frac{20x+15}{x^2+4x+3} = \frac{A.(x+3)}{(x+1)} + \frac{B.(x+1)}{(x+3)}

20x+15 = A.(x+3) + B.(x+1)

Putting x = -1
-20 + 15 = A(2) + B(0)

-5 = A\cdot2

A = \frac{-5}{2}

Putting x = -3
-60 + 15 = A(0) + B(-2)

-45 = -2 \cdot B

B = \frac{45}{2}

\frac{20x+15}{x^2+4x+3} = \frac{-5}{2(x+1)} + \frac{45}{2(x+3)}

I_2 = \int_{1}^{2} \frac{-5}{2(x+1)} + \frac{45}{2(x+3)} \ dx

I_2 = \frac{-5}{2} \int_{1}^{2} \frac{1}{(x+1)} \ dx + \frac{45}{2} \int_{1}^{2} \frac{1}{(x+3)} \ dx

I_2 = \frac{-5}{2} \Big[ \log(x+1) \Big]_{1}^{2} + \frac{45}{2} \Big [ \log(x+3) \Big]_{1}^{2}

I_2 = \frac{-5}{2} \Big[ \log{3} - log{2} \Big] + \frac{45}{2} \Big [ \log{5} - \log{4} \Big]

I_2 = \frac{-5}{2} \log{\frac{3}{2}}+ \frac{45}{2} \log{\frac {5}{4}}

I_2 = \frac{5}{2} \big[ 9\log{\frac {5}{4}} - \log{\frac{3}{2}} \Big]

I = I_1 - I_2

I = 5 - \frac{5}{2} \big[ 9\log{\frac {5}{4}} - \log{\frac{3}{2}} \Big]

Final Answer: \boxed{ \int_{1}^{2} \dfrac{5x^{2}}{x^{2}+4x+3} \ dx = 5 - \frac{5}{2} \big[ 9\log{\frac {5}{4}} - \log{\frac{3}{2}} \Big] }

Question 17: \int_{0}^{\tfrac{\pi}{4}} \big(2\sec^{2}x + x^{3} + 2\big) \ dx

Let, I = \int_{0}^{\tfrac{\pi}{4}} \big(2\sec^{2}x + x^{3} + 2\big) \ dx

I = \int_{0}^{\tfrac{\pi}{4}} 2\sec^{2}x \ dx + \int_{0}^{\tfrac{\pi}{4}} x^{3} \ dx + \int_{0}^{\tfrac{\pi}{4}} 2 \ dx

I = \Big[ 2\tan x + \frac{x^4}{4} + 2x]_{0}^{\frac{\pi}{4}}

I = \Big[ 2\tan \frac{\pi}{4} + \frac{(\frac{\pi}{4})^4}{4} + 2\cdot\frac{\pi}{4} \Big] - \Big[ 2\tan 0 + \frac{0^4}{4} + 2\cdot0 \Big]

I = \Big[ 2\cdot1 + \frac{\pi^4}{1024} + \frac{\pi}{2} \Big] - \Big[ 0 + 0 + 0 \Big]

Final Answer: \boxed{ \int_{0}^{\tfrac{\pi}{4}} \big(2\sec^{2}x + x^{3} + 2\big) \ dx = 2 + \frac{\pi^4}{1024} + \frac{\pi}{2}}

Question 18: \int_{0}^{\pi} \big(\sin^{2}\tfrac{x}{2} - \cos^{2}\tfrac{x}{2}\big) \ dx

Solution:
I = \int_{0}^{\pi} \big(\sin^{2}\tfrac{x}{2} - \cos^{2}\tfrac{x}{2}\big) \ dx

I = \int_{0}^{\pi} \big( -\cos2x \big) \ dx

I = \int_{0}^{\pi} (-\cos 2x) \ dx

I = \Big[\frac{-\sin 2x}{2} \Big]_{0}^{\pi}

I = \Big[\frac{-\sin 2\pi}{2} \Big] - \Big[\frac{-\sin 0}{2} \Big]

I = 0

Final Answer: \boxed {\int_{0}^{\pi} \big(\sin^{2}\tfrac{x}{2} - \cos^{2}\tfrac{x}{2}\big) \ dx = 0 }

Question 19: \int_{0}^{2} \dfrac{6x+3}{x^{2}+4} \ dx

Solution:
I = \int_{0}^{2} \dfrac{6x+3}{x^{2}+4} \ dx

I = \int_{0}^{2} \dfrac{6x}{x^{2}+4} \ dx + \int_{0}^{2} \dfrac{3}{x^{2}+4} \ dx

I = \int_{0}^{2} 3\cdot\dfrac{2x}{x^{2}+4} \ dx + \int_{0}^{2} \dfrac{3}{x^{2}+4} \ dx

I = I_1 + I_2

I_1 = \int_{0}^{2} 3\cdot\dfrac{2x}{x^{2}+4} \ dx

let x^2 + 4 = t \implies 2x \cdot dx = dt

\int 3\cdot\dfrac{2x}{x^{2}+4} \ dx = \int \frac{3 dt}{t} \ dx

\int 3\cdot\dfrac{2x}{x^{2}+4} \ dx = 3 \cdot \log{t}

Substituting back t
\int 3\cdot\dfrac{2x}{x^{2}+4} \ dx = 3 \cdot \log(x^2+4)

I_1 = 3 \Big[ \log(x^2 + 4) \Big]_{0}^{2}

I_1 = 3 \Big[ \log{8} - \log{4} \Big]

I_1 = 3 \log{2}

I_2 = \int_{0}^{2} \dfrac{3}{x^{2}+4} \ dx

\int \dfrac{3}{x^{2}+4} \ dx = 3 \cdot \frac{1}{2} \tan^{-1} \frac{x}{2}

I_2 = 3 \Big[\frac{1}{2} \tan^{-1} \frac{x}{2} \Big]_{0}^{2}

I_2 = \frac{3}{2} \Big[\tan^{-1} \frac{2}{2} - \tan^{-1} \frac{0}{2} \Big]

I_2 = \frac{3}{2} \Big[\tan^{-1} 1 - \tan^{-1} 0 \Big]

I_2 = \frac{3}{2} \Big[\frac{\pi}{4} - 0 \Big]

I_2 = \frac{3 \pi}{8}

I = 3 \log{2} + \frac{3 \pi}{8}

Final Answer: \boxed{ \int_{0}^{2} \dfrac{6x+3}{x^{2}+4} \ dx = 3 \log{2} + \frac{3 \pi}{8} }

Question 20: \int_{0}^{1} \big(x e^{x} + \sin \tfrac{\pi x}{4}\big), dx

I = \int_{0}^{1} \big(x e^{x} + \sin \tfrac{\pi x}{4}\big) \ dx

\int x e^{x} \ dx = x e^{x} - e^{x}

\int \sin \tfrac{\pi x}{4} \ dx = -\tfrac{4}{\pi}\cos \tfrac{\pi x}{4}

\therefore I= \Big[x e^{x} - e^{x}\Big]_{0}^{1} + \Big[-\tfrac{4}{\pi}\cos \tfrac{\pi x}{4}\Big]_{0}^{1}

I = \Big[1\cdot e^{1} - e^{1}\Big] - \Big[0\cdot e^{0} - e^{0}\Big] + \Big[-\tfrac{4}{\pi}\cos \tfrac{\pi}{4}\Big] - \Big[-\tfrac{4}{\pi}\cos \tfrac{0}{4}\Big]

I = 1 -\frac{2\sqrt{2}}{\pi} + \frac{4}{\pi}

Final Answer: \boxed{\int_{0}^{1} \big(x e^{x} + \sin \tfrac{\pi x}{4}\big) \ dx = 1 + \frac{4}{\pi} - \frac{2\sqrt{2}}{\pi} }

Question 21: \int_{1}^{\sqrt{3}} \dfrac{dx}{1+x^{2}}

(A) \frac {\pi}{3}

(B) \frac {2\pi}{3}

(C) \frac {\pi}{6}

(D) \frac {\pi}{12}

Solution:
I = \int_{1}^{\sqrt{3}} \dfrac{dx}{1+x^{2}}

I = \Big[ \tan^{-1} x \Big]_{1}^{\sqrt{3}}

I = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1)

I = \frac{\pi}{3} - \frac{\pi}{4}

I = \frac{\pi}{12}

\boxed{ \int_{1}^{\sqrt{3}} \dfrac{dx}{1+x^{2}} = \dfrac{\pi}{12}}

Correct option: (D)

Question 22: \int_{0}^{\tfrac{2}{3}} \dfrac{dx}{4+9x^{2}}

(A) \frac{\pi}{6}

(B) \frac{\pi}{12}

(C) \frac{\pi}{24}

(D) \frac{\pi}{4}

Solution:
I = \int_{0}^{\tfrac{2}{3}} \dfrac{dx}{4+9x^{2}}

I = \frac{1}{9} \int_{0}^{\tfrac{2}{3}} \dfrac{dx}{ \frac{4}{9}+9x^{2}}

I = \frac{1}{9} \int_{0}^{\tfrac{2}{3}} \dfrac{dx}{ (\frac{2}{3})^2+ x^{2}}

I = \frac{1}{9} \Big[ \frac{3}{2} \tan^{-1} \frac{3x}{2} \Big]_{0}^{\frac{2}{3}}

I = \frac{1}{6} \Big[ \tan^{-1} {1} - \tan^{-1}{0} \Big]

I = \frac{1}{6} \Big[ \frac{\pi}{4} - 0 \Big]

I = \dfrac{\pi}{24}

\boxed{\int_{0}^{\tfrac{2}{3}} \dfrac{dx}{4+9x^{2}} = \dfrac{\pi}{24} }

Correct Option: (C)