Exercise 7.9 Class 12 Maths Integrals

NCERT Solutions Class 12 Maths Chapter 7 Ex 7.9 | LearnCBSEHub

💡 Substitution in Definite Integrals

When you substitute $x = g(t)$, you must also change the limits of integration.

If lower limit is $x=a$, find new limit $t$ such that $a = g(t)$.
If upper limit is $x=b$, find new limit $t$ such that $b = g(t)$.
Advantage: You don’t need to substitute back to $x$ at the end. Just evaluate with new $t$ limits.

Question 1

$\int_0^1 \frac{x}{x^2+1} dx$

Step 1: Substitute

Let $t = x^2+1 \implies dt = 2x dx \implies x dx = dt/2$.
New Limits:
At $x=0 \implies t = 0^2+1 = 1$.
At $x=1 \implies t = 1^2+1 = 2$.

Step 2: Evaluate

$I = \int_1^2 \frac{1}{t} \frac{dt}{2} = \frac{1}{2} [\log|t|]_1^2$
$= \frac{1}{2} (\log 2 – \log 1) = \frac{1}{2}\log 2$.

Result: $\frac{1}{2}\log 2$

Question 2

$\int_0^{\pi/2} \sqrt{\sin\phi} \cos^5\phi d\phi$

Step 1: Simplify & Substitute

$\cos^5\phi = \cos^4\phi \cdot \cos\phi = (1-\sin^2\phi)^2 \cos\phi$.
Let $\sin\phi = t \implies \cos\phi d\phi = dt$.
Limits: $\phi=0 \to t=0$; $\phi=\pi/2 \to t=1$.

Step 2: Integrate

$I = \int_0^1 \sqrt{t} (1-t^2)^2 dt = \int_0^1 t^{1/2}(1 – 2t^2 + t^4) dt$
$= \int_0^1 (t^{1/2} – 2t^{5/2} + t^{9/2}) dt$
$= [\frac{2}{3}t^{3/2} – 2(\frac{2}{7})t^{7/2} + \frac{2}{11}t^{11/2}]_0^1$
$= \frac{2}{3} – \frac{4}{7} + \frac{2}{11} = \frac{154 – 132 + 42}{231} = \frac{64}{231}$.

Result: $\frac{64}{231}$

Question 3

$\int_0^1 \sin^{-1}(\frac{2x}{1+x^2}) dx$

Step 1: Trig Substitution

Let $x = \tan\theta \implies dx = \sec^2\theta d\theta$.
$\frac{2x}{1+x^2} = \sin 2\theta$. So integrand becomes $\sin^{-1}(\sin 2\theta) = 2\theta$.
Limits: $x=0 \to \theta=0$; $x=1 \to \theta=\pi/4$.

Step 2: Integration by Parts

$I = \int_0^{\pi/4} 2\theta \sec^2\theta d\theta$.
$= 2 [\theta \tan\theta – \int \tan\theta d\theta]_0^{\pi/4}$
$= 2 [\theta \tan\theta – \log|\sec\theta|]_0^{\pi/4}$
$= 2 [(\frac{\pi}{4}(1) – \log\sqrt{2}) – (0 – 0)] = \frac{\pi}{2} – 2(\frac{1}{2}\log 2) = \frac{\pi}{2} – \log 2$.

Result: $\frac{\pi}{2} – \log 2$

Question 4

$\int_0^2 x\sqrt{x+2} dx$ (Put $x+2=t^2$)

Step 1: Substitute

$x+2=t^2 \implies dx = 2t dt$. Also $x = t^2-2$.
Limits: $x=0 \to t=\sqrt{2}$; $x=2 \to t=\sqrt{4}=2$.

Step 2: Integrate

$I = \int_{\sqrt{2}}^2 (t^2-2) \cdot t \cdot 2t dt = 2 \int_{\sqrt{2}}^2 (t^4 – 2t^2) dt$
$= 2 [\frac{t^5}{5} – \frac{2t^3}{3}]_{\sqrt{2}}^2$
Upper Limit: $2(\frac{32}{5} – \frac{16}{3}) = 2(\frac{96-80}{15}) = \frac{32}{15}$.
Lower Limit: $2(\frac{4\sqrt{2}}{5} – \frac{4\sqrt{2}}{3}) = 8\sqrt{2}(\frac{3-5}{15}) = -\frac{16\sqrt{2}}{15}$.
Total: $\frac{16\sqrt{2} + 32}{15}$.

Result: $\frac{16}{15}(\sqrt{2} + 2)$

Question 5

$\int_0^{\pi/2} \frac{\sin x}{1+\cos^2 x} dx$

Step 1: Substitute

Let $\cos x = t \implies -\sin x dx = dt \implies \sin x dx = -dt$.
Limits: $x=0 \to t=1$; $x=\pi/2 \to t=0$.

Step 2: Evaluate

$I = \int_1^0 \frac{-dt}{1+t^2} = \int_0^1 \frac{dt}{1+t^2}$ (Swapping limits changes sign)
$= [\tan^{-1} t]_0^1 = \tan^{-1} 1 – \tan^{-1} 0 = \frac{\pi}{4}$.

Result: $\frac{\pi}{4}$

Question 6

$\int_0^2 \frac{dx}{x+4-x^2}$

Step 1: Complete Square

$4+x-x^2 = -(x^2-x-4) = -[(x-\frac{1}{2})^2 – \frac{1}{4} – 4] = (\frac{\sqrt{17}}{2})^2 – (x-\frac{1}{2})^2$.
Use $\int \frac{dx}{a^2-x^2} = \frac{1}{2a}\log|\frac{a+x}{a-x}|$.

Step 2: Evaluate

$I = [\frac{1}{\sqrt{17}} \log|\frac{\frac{\sqrt{17}}{2} + (x-1/2)}{\frac{\sqrt{17}}{2} – (x-1/2)}|]_0^2$
$= \frac{1}{\sqrt{17}} \log|\frac{\sqrt{17}+2x-1}{\sqrt{17}-2x+1}|_0^2$.
Upper ($x=2$): $\log|\frac{\sqrt{17}+3}{\sqrt{17}-3}|$. Lower ($x=0$): $\log|\frac{\sqrt{17}-1}{\sqrt{17}+1}|$.

Result: $\frac{1}{\sqrt{17}}\log(\frac{21+5\sqrt{17}}{4})$

Question 7

$\int_{-1}^1 \frac{dx}{x^2+2x+5}$

$x^2+2x+5 = (x+1)^2 + 2^2$.
$I = \int_{-1}^1 \frac{dx}{(x+1)^2+2^2} = [\frac{1}{2}\tan^{-1}(\frac{x+1}{2})]_{-1}^1$.
$= \frac{1}{2} [\tan^{-1}(1) – \tan^{-1}(0)] = \frac{1}{2}(\frac{\pi}{4}) = \frac{\pi}{8}$.

Result: $\frac{\pi}{8}$

Question 8

$\int_1^2 (\frac{1}{x} – \frac{1}{2x^2}) e^{2x} dx$

Step 1: Substitute

Let $2x = t \implies dx = dt/2$. Limits: $1 \to 2$, $2 \to 4$.
Integrand: $(\frac{2}{t} – \frac{1}{2(t/2)^2}) e^t \frac{dt}{2} = \frac{1}{2} (\frac{2}{t} – \frac{2}{t^2}) e^t dt = (\frac{1}{t} – \frac{1}{t^2}) e^t dt$.

Step 2: Property $\int e^t(f+f’)dt$

Here $f(t) = 1/t, f'(t) = -1/t^2$.
$I = [\frac{e^t}{t}]_2^4 = \frac{e^4}{4} – \frac{e^2}{2} = \frac{e^2(e^2-2)}{4}$.

Result: $\frac{e^2(e^2-2)}{4}$

Questions 9 — 10

MCQs

9. $\int_{1/3}^1 \frac{(x-x^3)^{1/3}}{x^4} dx$

Numerator: $(x^3(\frac{1}{x^2}-1))^{1/3} = x (\frac{1}{x^2}-1)^{1/3}$.
Integrand: $\frac{1}{x^3}(\frac{1}{x^2}-1)^{1/3}$.
Let $\frac{1}{x^2}-1 = t \implies -\frac{2}{x^3}dx = dt \implies \frac{dx}{x^3} = -dt/2$.
Limits: $x=1/3 \to t=8$; $x=1 \to t=0$.
$I = \int_8^0 t^{1/3} (-dt/2) = \frac{1}{2}\int_0^8 t^{1/3} dt = \frac{1}{2}[\frac{3}{4}t^{4/3}]_0^8$
$= \frac{3}{8}(16 – 0) = 6$.

Correct Option: (A) 6

10. $f(x) = \int_0^x t \sin t dt$, find $f'(x)$

Using Fundamental Theorem of Calculus: $\frac{d}{dx} \int_0^x g(t) dt = g(x)$.
Here $g(t) = t \sin t$. So $f'(x) = x \sin x$.

Correct Option: (B) $x \sin x$