Class 12 Integration Exercise 7.8 NCERT Solutions (Definite Integration)

If you are looking for Class 12 NCERT integration solutions you have come to the right place.

Integration is one of the most important topics in Class 12 Maths. Here you will find NCERT Solutions for Class 12 Maths Integration Exercise 7.8 (based on definite integration), solved step by step. These solutions follow the CBSE exam style and help you understand how to apply limits.

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)