Exercise 11.12 - Chapter 11 Integral Calculus 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 11.12
Integrate the following with respect to $\mathrm{x}$ :
Question 1.
(i) $\sqrt{x^2+2 x+10}$
Solution:
$
\begin{aligned}
& =\int \sqrt{(x+1)^2-1+10} d x \\
& =\int \sqrt{(x+1)^2+(3)^2} d x \\
& =\frac{(x+1)}{2} \sqrt{(x+1)^2+(3)^2}+\frac{(3)^2}{2} \log \left|(x+1)+\sqrt{(x+1)^2}\right| \\
& =\frac{(x+1)}{2} \sqrt{x^2+2 x+10}+\frac{9}{2} \log \left|(x+1)+\sqrt{x^2+2 x+10}\right|+c
\end{aligned}
$
(ii) $\sqrt{x^2-2 x-3}$
Solution:
$
\begin{aligned}
& =\int \sqrt{(x-1)^2-1-3} d x \\
& =\int \sqrt{(x-1)^2-(2)^2} d x \\
& =\frac{(x-1)}{2} \sqrt{(x-1)^2-(2)^2}-\frac{(2)^2}{2} \log \left|(x-1)+\sqrt{(x-1)^2-(2)^2}\right|+c \\
& =\frac{(x-1)}{2} \sqrt{x^2-2 x-3}-2 \log \left|(x-1)+\sqrt{x^2-2 x-3}\right|+c
\end{aligned}
$
(iii) $\sqrt{(6-x)(x-4)}$
Solution:

$\begin{aligned}
& =\int \sqrt{10 x-x^2-24} d x \\
& =-\left[x^2-10 x+24\right] \\
& =-\left[(x-5)^2-25+24\right] \\
& =-\left[(x-5)^2-(1)^2\right] \\
& =1-(x-5)^2 \\
& =\int \sqrt{1-(x-5)^2} d x \\
& =\frac{x-5}{2} \sqrt{1-(x-5)^2}+\frac{1}{2} \sin ^{-1}\left(\frac{x-5}{1}\right) \\
& =\frac{x-5}{2} \sqrt{10 x-x^2-24}+\frac{1}{2} \sin ^{-1}(x-5)+c
\end{aligned}$

Question 2.
(i) $\sqrt{9-(2 x+5)^2}$
Solution:
$
\begin{gathered}
=\int \sqrt{(3)^2-(2 x+5)^2} d x \\
=\frac{1}{2}\left[\left(\frac{2 x+5}{2}\right) \sqrt{9-(2 x+5)^2}+\frac{9}{2} \sin ^{-1}\left(\frac{2 x+5}{3}\right)\right]+c \\
=\frac{1}{4}\left[\left[(2 x+5) \sqrt{9-(2 x+5)^2}+9 \sin ^{-1}\left(\frac{2 x+5}{3}\right)\right]+c\right.
\end{gathered}
$
(ii) $\sqrt{81+(2 x+1)^2}$
Solution:
$
\begin{aligned}
& =\int \sqrt{(9)^2+(2 x+1)^2} d x \\
& =\frac{1}{2}\left[\left(\frac{2 x+1}{2}\right) \sqrt{81+(2 x+1)^2}+\frac{81}{2} \log \left|(2 x+1)+\sqrt{81+(2 x+1)^2}\right|\right]+c \\
& =\frac{1}{4}\left[(2 x+1) \sqrt{81+(2 x+1)^2}+81 \log \left|(2 x+1)+\sqrt{81+(2 x+1)^2}\right|\right]+c
\end{aligned}
$
(iii) $\sqrt{(x+1)^2-4}$
Solution:

$\begin{aligned}
& =\int \sqrt{(x+1)^2-(2)^2} d x \\
& =\left(\frac{x+1}{2}\right) \sqrt{(x+1)^2-4}-\frac{(2)^2}{2} \log \left|(x+1)+\sqrt{(x+1)^2-4}\right|+c \\
& =\left(\frac{x+1}{2}\right) \sqrt{(x+1)^2-4}+2 \log \mid(x+1)+\sqrt{(x+1)^2-4}+c
\end{aligned}$