Class 12 NCERT Maths – Chapter 7, Exercise 7.9 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 7.9 (based on definite integration), solved step by step. These solutions follow the CBSE exam style and help you understand how to apply limits both with and without substitution methods.

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_{0}^{1} \dfrac{x}{x^{2}+1} \ dx

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

Let t = x^{2} + 1 \implies dt = 2x dx
So, x dx = \dfrac{dt}{2}.

Limits: when x=0 \implies t=1, when x=1 \implies t=2.

I = \int_{1}^{2} \dfrac{1}{t} \cdot \dfrac{dt}{2}

I = \dfrac{1}{2} \int_{1}^{2} \dfrac{1}{t} \ dt

I = \dfrac{1}{2} [\ln t]_{1}^{2}

I = \dfrac{1}{2} [\ln 2 - \ln 1]

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

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

Question 2: Evaluate \int_{0}^{\pi/2} \sqrt{\sin \phi} \ \cos^{5}\phi \ d\phi

Solution:
I = \int_{0}^{\pi/2} \sqrt{\sin \phi}, \cos^{5}\phi \ d\phi

Let t = \sin \phi \implies dt = \cos \phi \ d\phi.

So,
\cos^{5}\phi \ d\phi = \cos^{4}\phi \cdot \cos\phi \ d\phi = (1 - \sin^{2}\phi)^{2} \ dt = (1 - t^{2})^{2} \ dt.

Limits: when \phi = 0 \implies t=0, when \phi = \dfrac{\pi}{2} \implies t=1.

I = \int_{0}^{\pi/2} \sqrt{\sin \phi} \cos^{5}\phi \ d\phi

I = \int_{0}^{1} \sqrt{t},(1 - t^{2})^{2} \ dt

I = \int_{0}^{1} (t^{1/2} - 2t^{5/2} + t^{9/2}) \ dt

Now integrate term by term:
I = \left[\dfrac{2}{3}t^{3/2} - \dfrac{4}{7}t^{7/2} + \dfrac{2}{11}t^{11/2}\right]_{0}^{1}

I = \dfrac{2}{3} - \dfrac{4}{7} + \dfrac{2}{11}

LCM (3,7,11) = 231

I = \dfrac{154 - 132 + 42}{231} = \dfrac{64}{231}

Final Answer: \boxed{\int_{0}^{\pi/2} \sqrt{\sin \phi} \ \cos^{5}\phi \ d\phi = \dfrac{64}{231}}

Question 3: Evaluate \int_{0}^{1} \sin^{-1}\left(\dfrac{2x}{1+x^{2}}\right) dx

Solution:
I = \int_{0}^{1} \sin^{-1}\left(\dfrac{2x}{1+x^{2}}\right) dx

Use substitution x = \tan \theta \implies dx = \sec^{2}\theta d\theta

Limits: when x = 0 \implies \theta =0, when x = 1 \implies \theta=\frac{\pi}{4}.

Then, \dfrac{2x}{1+x^{2}} = \dfrac{2\tan\theta}{1+\tan^{2}\theta} = \sin 2\theta

So integral becomes:

I = \int_{0}^{\pi/4} \sin^{-1}(\sin 2\theta) \ \sec^{2}\theta d\theta

Since 0 \leq 2\theta \leq \pi/2, we get \sin^{-1}(\sin 2\theta) = 2\theta.

So,
I = \int_{0}^{\pi/4} 2\theta \sec^{2}\theta \ d\theta.

Use integration by parts:
Let u = 2\theta, \ v = \sec^{2}\theta.
Then du = 2, d\theta, \int v \ d\theta = \tan \theta.

I = [2\theta \cdot \tan\theta]_{0}^{\pi/4} - \int_{0}^{\pi/4} 2 \tan \theta, d\theta

I = [2\theta \cdot \tan\theta]_{0}^{\pi/4} - 2[\ln(\sec \theta)]_{0}^{\pi/4}

I = \Big(\frac{\pi}{2} \cdot 1\Big) - 2 \Big [\ln(\sec \frac{\pi}{4}) - \ln(\sec 0) \Big]

I = \dfrac{\pi}{2} - 2\left(\ln\left(\sqrt{2}\right) + 0\right)

I = \dfrac{\pi}{2} - 2\ln \sqrt{2}

I = \dfrac{\pi}{2} - \ln 2

Final Answer: \boxed{\int_{0}^{1} \sin^{-1}\left(\dfrac{2x}{1+x^{2}}\right) dx = \dfrac{\pi}{2} - \ln 2}

Question 4: Evaluate \int_{0}^{2} x \sqrt{x+2} \ dx

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

Let t = x+2 \implies dt = dx, and x = t-2
Limits: when x=0 \implies t=2, when x=2 \implies t=4

I = \int_{2}^{4} (t-2)\sqrt{t} \ dt

I = \int_{2}^{4} (t^{3/2} - 2t^{1/2}), dt

I = \left[\dfrac{2}{5}t^{5/2} - \dfrac{4}{3}t^{3/2}\right]_{2}^{4}

At t=4: \dfrac{2}{5}(32) - \dfrac{4}{3}(8) = \dfrac{64}{5} - \dfrac{32}{3}

At t=2: \dfrac{2}{5}(2^{5/2}) - \dfrac{4}{3}(2^{3/2}) = \dfrac{2}{5}(4\sqrt{2}) - \dfrac{4}{3}(2\sqrt{2}) = \dfrac{8\sqrt{2}}{5} - \dfrac{8\sqrt{2}}{3}

So, I = \left(\dfrac{64}{5} - \dfrac{32}{3}\right) - \left(\dfrac{8\sqrt{2}}{5} - \dfrac{8\sqrt{2}}{3}\right)

I = \dfrac{192 - 160}{15} - \dfrac{24\sqrt{2} - 40\sqrt{2}}{15}

I = \dfrac{32}{15} + \dfrac{16\sqrt{2}}{15}

Final Answer: \boxed{\int_{0}^{2} x \sqrt{x+2} \ dx = \dfrac{16\sqrt{2}}{15} \Big( \sqrt{2} + 1\Big)}

Question 5: Evaluate \int_{0}^{\pi/2} \dfrac{\sin x}{1 + \cos^{2}x} \ dx

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

Let t = \cos x \implies dt = -\sin x, dx.
Limits: when x=0 \implies t=1, when x=\pi/2 \implies t=0.

I = \int_{0}^{\pi/2} \dfrac{\sin x}{1 + \cos^{2}x} \ dx

I = \int_{1}^{0} \dfrac{-dt}{1+t^{2}}

I = \int_{0}^{1} \dfrac{dt}{1+t^{2}}

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

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

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

Question 6: Evaluate \int_{0}^{2}\dfrac{dx}{x+4-x^{2}}

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

I = -\int_{0}^{2} \dfrac{dx}{x^{2}-x-4}

Completing the square:
x^{2}-x-4 = (x-\tfrac{1}{2})^{2}-\left(\tfrac{\sqrt{17}}{2}\right)^{2}

So,
I = -\int_{0}^{2} \dfrac{dx}{(x-\tfrac{1}{2})^{2}-\left(\tfrac{\sqrt{17}}{2}\right)^{2}}

Using standard formula

I = -\dfrac{1}{2 \tfrac{\sqrt{17}}{2}} \left[\ln \left|\dfrac{x-\tfrac{1}{2}-\tfrac{\sqrt{17}}{2}}{x-\tfrac{1}{2}+\tfrac{\sqrt{17}}{2}}\right|\right]_{0}^{2}

I = -\dfrac{1}{2 \tfrac{\sqrt{17}}{2}} \left[\ln \left|\dfrac{x-\tfrac{1}{2}-\tfrac{\sqrt{17}}{2}}{x-\tfrac{1}{2}+\tfrac{\sqrt{17}}{2}}\right|\right]_{0}^{2}

I = - \dfrac{1}{\sqrt{17}} \Big( \left[\ln \left|\dfrac{2-\tfrac{1}{2}-\tfrac{\sqrt{17}}{2}}{2-\tfrac{1}{2}+\tfrac{\sqrt{17}}{2}}\right|\right] - \left[\ln \left|\dfrac{0-\tfrac{1}{2}-\tfrac{\sqrt{17}}{2}}{0-\tfrac{1}{2}+\tfrac{\sqrt{17}}{2}}\right|\right] \big)

I = - \dfrac{1}{\sqrt{17}} \Big( \left[\ln \left|\dfrac{3 - \sqrt{17}}{3+ \sqrt{17}} \right|\right] - \left[\ln \left|\dfrac{-1-\sqrt{17}}{-1+\sqrt{17}}\right|\right] \big)

I = - \dfrac{1}{\sqrt{17}} \Big( \left[\ln \left|\dfrac{3 - \sqrt{17}}{3+ \sqrt{17}} \right|\right] - \left[\ln \left|\dfrac{1+\sqrt{17}}{1-\sqrt{17}}\right|\right] \big)

I = - \dfrac{1}{\sqrt{17}} \left[\ln \left|\dfrac{(3 - \sqrt{17})(1 - \sqrt{17})}{(3+ \sqrt{17})(1 + \sqrt{17})} \right|\right]

I = - \dfrac{1}{\sqrt{17}} \left[\ln \left|\dfrac{20 - 4\sqrt{17}}{20 + 4\sqrt{17}} \right|\right]

I = \dfrac{1}{\sqrt{17}} \left[\ln \left|\dfrac{20 + 4\sqrt{17}}{20 - 4\sqrt{17}} \right|\right]

I = \dfrac{1}{\sqrt{17}} \left[\ln \left|\dfrac{5 + \sqrt{17}}{5 - \sqrt{17}} \right|\right]

Multiplying and dividing by 5 + \sqrt{17}
I = \dfrac{1}{\sqrt{17}} \left[\ln \left|\dfrac{(5 + \sqrt{17})(5 + \sqrt{17})}{(5 - \sqrt{17})(5 + \sqrt{17})} \right|\right]

I = \dfrac{1}{\sqrt{17}} \left[\ln \left|\dfrac{25 + 10\sqrt{17} + 17}{25 - 17} \right|\right]

I = \dfrac{1}{\sqrt{17}} \left[\ln \left|\dfrac{42 + 10\sqrt{17}}{8} \right|\right]

I = \dfrac{1}{\sqrt{17}} \left[\ln \left|\dfrac{21 + 5\sqrt{17}}{4} \right|\right]

Final Answer: \boxed{\int_{0}^{2} \dfrac{dx}{x+4-x^{2}} = \dfrac{1}{\sqrt{17}} \left[\ln \left|\dfrac{21 + 5\sqrt{17}}{4} \right|\right] }

Question 7: Evaluate \int_{-1}^{1}\dfrac{dx}{x^{2}+2x+5}

Solution:
I = \int_{-1}^{1} \dfrac{dx}{x^{2}+2x+5}

Completing the square: x^{2}+2x+5 = (x+1)^{2}+2^{2}

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

Let t = x+1 \implies dx = dt.
Limits: when x=-1, t=0; when x=1, t=2.

I = \int_{0}^{2} \dfrac{dt}{t^{2}+4}

I = \left[\dfrac{1}{2}\tan^{-1}\left(\dfrac{t}{2}\right)\right]_{0}^{2}

I = \dfrac{1}{2} \left[\tan^{-1}1 - \tan^{-1}0 \right]

I = \dfrac{1}{2}\left(\dfrac{\pi}{4}-0\right)

I = \dfrac{\pi}{8}

Final Answer: \boxed{\int_{-1}^{1} \dfrac{dx}{x^{2}+2x+5} = \dfrac{\pi}{8}}

Question 8: Evaluate \int_{1}^{2}\left(\dfrac{1}{x}-\dfrac{1}{2x^{2}}\right)e^{2x} dx

Solution:
I = \int_{1}^{2}\left(\dfrac{1}{x}-\dfrac{1}{2x^{2}}\right)e^{2x} dx

Let t = 2x \implies dt = 2 dx \implies \frac{dt}{2} = dx

Limits when x = 1, t = 2 ; when x = 2, t = 4

I = \int_{2}^{4}\left(\dfrac{2}{t}-\dfrac{2}{t^{2}}\right)e^{t} \ \frac{dt}{2}

I = \int_{2}^{4}\left(\dfrac{1}{t}-\dfrac{1}{t^{2}}\right)e^{t} \ dt

Here f(x) = \dfrac{1}{t} \ \text{and} \ f'(x) = \dfrac{-1}{t^2}

We know, \int e^x [f(x) + f'(x)] dx = e^x f(x) + C

I = \int_{2}^{4}\left(\dfrac{1}{t} + \Big(-\dfrac{1}{t^{2}} \Big) \right) e^{t} \ dt

I = \Big[e^{t} \cdot \dfrac{1}{t} \Big]_{2}^{4}

I = \Big[\frac{e^{4}}{4} - \frac{e^{2}}{2} \Big]

I = \dfrac{e^2(e^2 - 2)}{4}

Final Answer: \boxed{ \int_{1}^{2}\left(\dfrac{1}{x}-\dfrac{1}{2x^{2}}\right)e^{2x} dx = \dfrac{e^2(e^2 - 2)}{4} }

Question 9: Find the value of \int_{1/3}^{1}\dfrac{(x-x^{3})^{1/3}}{x^{4}} \ dx

(A) 6

(B) 0

(C) 3

(D) 4

Solution:
I = \int_{1/3}^{1}\dfrac{(x-x^{3})^{1/3}}{x^{4}} \ dx

I = \int_{1/3}^{1}\dfrac{(x^3(\tfrac{1}{x^2}-1))^{1/3}}{x^{4}} dx

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

Let t = \dfrac{1}{x^{2}}-1 \implies dt = -\dfrac{2}{x^{3}}dx.

\dfrac{-dt}{2} = \dfrac{dx}{x^3}

Limits when x = \frac{1}{3}, t = 8 ; when x = 1, t = 0

So, I = \int_{8}^{0} t^{1/3}\cdot \dfrac{-dx}{2}

I = \frac{1}{2} \int_{8}^{0} - t^{1/3} \ dt

I = \dfrac{1}{2}\int_{0}^{8} t^{1/3} \ dt

I = \dfrac{1}{2} \Big[ \dfrac{t^{\tfrac{4}{3}}}{\tfrac{4}{3}}\Big]_{0}^{8}

I = \dfrac{3}{8} \Big[ 8^{\tfrac{4}{3}} - 0^{\tfrac{4}{3}} \Big]

I = \dfrac{3}{8} \Big[ 16 \Big]

I = 6

\boxed{ \int_{1/3}^{1}\dfrac{(x-x^{3})^{1/3}}{x^{4}} \ dx = 6}

Correct Option: (A)

Question 10: If f(x)=\int_{0}^{x} t\sin t \ dt, then find f'(x)

(A) \cos x + x \sin x

(B) x \sin x

(C) x \cos x

(D) \sin x + x \cos x

Solution:
f(x)=\int_{0}^{x} t\sin t \ dt

Differentiating,
f'(x) = \dfrac {d}{dx} \int_{0}^{x} t\sin t \ dt

f'(x) = \Big[ t \sin t \Big]_{0}^{x}

f'(x) = x \sin x - 0

f'(x) = x\sin x

\boxed{f'(x) = x\sin x}

Correct Option: (B)