Exercise 11.10 - Chapter 11 Integral Calculus 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 11.10
Find the integrals of the following

Question 1.
(i) $\frac{1}{4-x^2}$
(ii) $\frac{1}{25-4 x^2}$
(iii) $\frac{1}{9 x^2-4}$
Solution:
(i)
$
\begin{aligned}
\int \frac{1}{4-x^2} d x & =\int \frac{1}{(2)^2-(x)^2} d x \\
\int \frac{1}{a^2-x^2} d x & =\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c \\
& =\frac{1}{2(2)} \log \left|\frac{2+x}{2-x}\right|+c \\
& =\frac{1}{4} \log \left|\frac{2+x}{2-x}\right|+c
\end{aligned}
$
(ii)
$
\begin{aligned}
\int \frac{1}{25-4 x^2} d x & =\int \frac{1}{(5)^2-(2 x)^2} d x \\
& =\frac{1}{2}\left[\frac{1}{2(5)} \log \left|\frac{5+2 x}{5-2 x}\right|\right]+c \\
& =\frac{1}{20} \log \left|\frac{5+2 x}{5-2 x}\right|+c
\end{aligned}
$
(iii)
$
\begin{aligned}
\int \frac{1}{9 x^2-4} d x & =\int \frac{1}{(3 x)^2-(2)^2} d x \\
\int \frac{1}{x^2-a^2} d x & =\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c \\
& =\frac{1}{3}\left\{\frac{1}{2(2)} \log \left|\frac{3 x-2}{3 x+2}\right|\right\}+c
\end{aligned}
$

$
=\frac{1}{12} \log \left|\frac{3 x-2}{3 x+2}\right|+c
$
Question 2.
(i) $\frac{1}{6 x-7-x^2}$
(ii) $\frac{1}{(x+1)^2-25}$
(iii) $\frac{1}{\sqrt{x^2+4 x+2}}$
Solution:

(i) Let $\mathrm{I}=\int \frac{1}{6 x-7-x^2} d x$
Consider, $-x^2+6 x-7=-\left[x^2-6 x+4\right]$
$
=-\left[(\mathrm{x}-3)^2-9+7\right]
$
$
\begin{aligned}
& =-\left[(x-3)^2-2\right] \\
& =-\left[(x-3)^2-(\sqrt{2})^2\right] \\
& =(\sqrt{2})^2-(x-3)^2 \\
I & =\int \frac{1}{(\sqrt{2})^2-(x-3)^2} d x \\
I & =\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2}+x-3}{\sqrt{2}-x+3}\right|+c
\end{aligned}
$
(ii)
$
\begin{aligned}
I=\int \frac{1}{(x+1)^2-25} d x & =\int \frac{1}{(x+1)^2-(5)^2} d x \\
& =\frac{1}{2(5)} \log \left|\frac{x+1-5}{x+1+5}\right|+c \\
& =\frac{1}{10} \log \left|\frac{x-4}{x+6}\right|+c
\end{aligned}
$

(iii) Let
$
I=\int \frac{1}{\sqrt{x^2+4 x+2}} d x
$
Consider,
$
\begin{aligned}
x^2+4 x+2 & =(x+2)^2-4+2 \\
& =(x+2)^2-2 \\
& =(x+2)^2-(\sqrt{2})^2 \\
\mathrm{I} & =\int \frac{1}{\sqrt{(x+2)^2-(\sqrt{2})^2}} \\
\int \frac{1}{\sqrt{x^2-a^2}} d x & =\log \left|x+\sqrt{x^2-a^2}\right|+c \\
& =\log |(x+2)| \sqrt{(x+2)^2-(\sqrt{2})^2} \mid+c \\
& =\log \left|(x+2)+\sqrt{x^2+4 x+2}\right|+c
\end{aligned}
$
Question 3.
(i) $\frac{1}{\sqrt{(2+x)^2-1}}$
(ii) $\frac{1}{\sqrt{x^2-4 x+5}}$
(iii) $\frac{1}{\sqrt{9+8 x-x^2}}$

Solution:

(i)
$
\begin{aligned}
\int \frac{1}{\sqrt{(2+x)^2-1}} d x & =\int \frac{1}{\sqrt{(2+x)^2-(1)^2}} d x \\
& =\log \left|(2+x)+\sqrt{(2+x)^2-1}\right|+c
\end{aligned}
$
(ii) Let
$
I=\int \frac{1}{\sqrt{x^2-4 x+5}} d x
$
Consider,
$
\begin{aligned}
& x^2-4 x+5=(x-2)^2-4+5 \\
&=(x-2)^2+(1)^2 \\
& \mathrm{I}=\int \frac{1}{\sqrt{(x-2)^2+(1)^2}} d x \\
& \int \frac{1}{\sqrt{x^2+a^2}} d x=\log \left|x+\sqrt{x^2+a^2}\right|+c
\end{aligned}
$
$
\begin{aligned}
& =\log \left|(x-2)+\sqrt{(x-2)^2+(1)^2}\right|+c \\
& =\log \left|(x-2)+\sqrt{x^2-4 x+5}\right|+c
\end{aligned}
$
(iii) Let
$
\mathrm{I}=\int \frac{1}{\sqrt{9+8 x-x^2}} d x
$
Consider, $\quad 9+8 x-x^2$
$
\begin{aligned}
& =-\left[x^2-8 x-9\right] \\
& =-\left[(x-4)^2-16-9\right] \\
& =-\left[(x-4)^2-(5)^2\right] \\
& =(5)^2-(x-4)^2 \\
I & =\int \frac{1}{\sqrt{(5)^2-(x-4)^2}} d x
\end{aligned}
$
$
\int \frac{1}{\sqrt{a^2-x^2}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c
$

$=\sin ^{-1}\left(\frac{x-4}{5}\right)+c$