Class 12 NCERT Maths – Chapter 7, Exercise 7.9 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_{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)