Exercise 9.3-Additional Problems - Chapter 9 Applications of Integration 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
Evaluate: $\int_{-\pi / 4}^{\pi / 4} x^3 \sin ^2 x d x$
Solution:
Let $f(x)=x^3 \sin ^2 x=x^3(\sin x)^2$
$
\begin{aligned}
& \therefore \mathrm{f}(-\mathrm{x})=(-\mathrm{x})^3(\sin (-\mathrm{x}))^2=(-\mathrm{x})^3(-\sin \mathrm{x})^2 \\
& =-\mathrm{x}^3 \sin ^2 \mathrm{x}=-\mathrm{f}(\mathrm{x}) \\
& \mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{x}) \text { is an odd function. } \\
& \therefore \int_{-\pi / 4}^{\pi / 4} x^3 \sin ^2 x d x=0 \text { (by property). }
\end{aligned}
$
Question 2.
Evaluate: $\int_{-1}^1 \log \left(\frac{3-x}{3+x}\right) d x$
Solution:
$
\begin{aligned}
\text { Let } f(x) & =\log \left(\frac{3-x}{3+x}\right) \\
\therefore f(-x) & =\log \left(\frac{3+x}{3-x}\right)=\log (3+x)-\log (3-x) \\
& =-[\log (3-x)-\log (3+x)] \\
& =-\left[\log \left(\frac{3-x}{3+x}\right)\right]=-f(x)
\end{aligned}
$
Thus $f(-x)=-f(x) \quad \therefore f(x)$ is an odd function.
$
\therefore \int_{-1}^1 \log \left(\frac{3-x}{3+x}\right) d x=0 .
$

Question 3.
Evaluate: $\int_{-\pi / 2}^{\pi / 2} x \sin x d x$
Solution:
Let $f(x)=x \sin x$
$f(-x)=(-x) \sin (-x)$
$=x \sin x(\because \sin (-x)=-\sin x)$
$\therefore \mathrm{f}(\mathrm{x})$ is an even function
$
\begin{aligned}
\int_{-\pi / 2}^{\pi / 2} x \sin x d x & =2 \int_0^{\pi / 2} x \sin x d x \\
& =2\left[\{x(-\cos x)\}_0^{\pi / 2}-\int_0^{\pi / 2}(-\cos x) d x\right]
\end{aligned}
$
Using the method of integration by parts
$
\begin{aligned}
& =2\left[0+\int_0^{\pi / 2} \cos x d x\right]=2[\sin x]_0^{\pi / 2} \\
& =2[1-0]=2 .
\end{aligned}
$
Question 4.
Evaluate: $\int_0^1 x(1-x)^n d x$

Solution:
$
\begin{aligned}
\text { Let } \mathrm{I} & =\int_0^1 x(1-x)^n d x \\
& =\int_0^1(1-x)[1-(1-x)]^n d x \quad\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right] \\
& =\int_0^1(1-x) x^n d x=\int_0^1\left(x^n-x^{n+1}\right) d x \\
& =\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_0^1=\left[\frac{1}{n+1}-\frac{1}{n+2}\right]=\frac{n+2-(n+1)}{(n+1)(n+2)} \\
\int_0^1 x(1-x)^n d x & =\frac{1}{(n+1)(n+2)}
\end{aligned}
$
Question 5.
Evaluate: $\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\cot x}}$
Solution:
Let $I=\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\cot x}}$
$
I=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x
$

$\begin{aligned}
& =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)} d x}{\sqrt{\sin \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}+\sqrt{\cos \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}} \\
& \left(\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right) \\
& I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x \\
& I=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \\
& \text { (1) }+(2) \Rightarrow 2 \mathrm{I}=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \\
& 2 \mathrm{I}=\int_{\pi / 6}^{\pi / 3} d x=[x]_{\pi / 6}^{\pi / 6}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6} \\
& \therefore \mathrm{I}=\frac{\pi}{12} \text {. } \\
&
\end{aligned}$