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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
$
\begin{aligned}
& \sqrt{(x+1)^2+4} \\
& \text { Solution: } \\
& \int \sqrt{(x+1)^2+4} d x \\
& =\frac{x+1}{2} \sqrt{4+(x+1)^2}+\frac{4}{2} \log (x+1)+\sqrt{(x+1)^2+4} \\
& =\frac{x+1}{2} \sqrt{4+(x+1)^2}+2 \log \left\{(x+1)+\sqrt{(x+1)^2+4}\right\}+c
\end{aligned}
$
Question 2.
$
\sqrt{(2 x+1)^2+9}
$
Solution:
$
\begin{aligned}
& \int \sqrt{(2 x+1)^2+9} d x=\int \sqrt{(2 x+1)^2+3^2} d x \\
& =\frac{\frac{1}{2}\left\{(2 x+1) \sqrt{(2 x+1)^2+3^2}\right\}+\frac{9}{2} \log (2 x+1)+\sqrt{(2 x+1)^2+9}}{2} \\
& =\frac{1}{4}\left\{(2 x+1) \sqrt{(2 x+1)^2+9}+9 \log \left\{(2 x+1)+\sqrt{(2 x+1)^2+9}\right\} c\right.
\end{aligned}
$
Question 3.
$
\sqrt{x^2-3 x+10}
$

Solution:
$
\begin{aligned}
x^2-3 x+10= & x^2-2\left(\frac{3}{2}\right) x+10+\frac{9}{4}-\frac{9}{4} \\
= & \left(x-\frac{3}{2}\right)^2+\frac{31}{4}=\left[\left(x-\frac{3}{2}\right)^2+\frac{\sqrt{31}}{2}\right]^2 \\
\therefore \quad \int \sqrt{x^2-3 x+10} d x= & \int \sqrt{\left(x-\frac{3}{2}\right)^2+\left(\frac{\sqrt{31}}{2}\right)^2} d x \\
= & \frac{1}{2}\left\{\left(x-\frac{3}{2}\right) \sqrt{\left(x-\frac{3}{2}\right)^2+\left(\frac{\sqrt{31}}{2}\right)^2}\right\} \\
& +\frac{31}{4} \log \left[\left(x-\frac{3}{2}\right)+\sqrt{\left(x-\frac{3}{2}\right)^2+\left(\frac{\sqrt{31}}{2}\right)^2}\right] \\
=\frac{2 x-3}{4} \sqrt{x^2-3 x+10}+ & \frac{31}{4} \log \left[\left(x-\frac{3}{2}\right)+\sqrt{x^2-3 x+10}\right]+c
\end{aligned}
$
Question 4.
$
\begin{aligned}
& \sqrt{169-(3 x+1)^2} \\
& \text { Solution: } \\
& \int \sqrt{169-(3 x+1)^2} d x \\
& =\frac{\frac{(3 x+1)}{2} \sqrt{169-(3 x+1)^2}+\frac{169}{2} \sin ^{-1}\left(\frac{3 x+1}{13}\right)}{3}+c \\
& =\frac{1}{6}\left\{3 x+1 \sqrt{169-(3 x+1)^2}+169 \sin ^{-1}\left(\frac{3 x+1}{13}\right)\right\}+c
\end{aligned}
$
Question 5 .
$
\sqrt{1-3 x-x^2}
$

Solution:
$
\begin{aligned}
& 1-3 x-x^2=-\left(x^2+3 x-1\right) \\
& =-\left[x^2+2\left(\frac{3}{2}\right) x-1+\frac{9}{4}-\frac{9}{4}\right] \\
& =-\left\{\left(x+\frac{3}{2}\right)^2-\frac{13}{4}\right\}=\frac{13}{4}-\left(x-\frac{3}{2}\right)^2 \\
& \therefore \quad \int \sqrt{1-3 x-x^2} d x=\int \sqrt{\frac{13}{4}-\left(x-\frac{3}{2}\right)^2} d x \\
& =\frac{\left(x-\frac{3}{2}\right)}{2} \sqrt{\frac{13}{4}-\left(x-\frac{3}{2}\right)^2}+\frac{\frac{13}{2}}{2} \sin ^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}}\right) \\
& =\frac{\left(x-\frac{3}{2}\right)}{2} \sqrt{1-3 x-x^2}+\frac{13}{8} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{13}}\right) \\
& =\frac{2 x-3}{4} \sqrt{1-3 x-x^2}+\frac{13}{8} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{13}}\right)+c \\
&
\end{aligned}
$
Question 6.
$
\sqrt{(2-x)(3+x)}
$
Solution:

$\begin{aligned}
& (2-x)(3+x)=6-x-x^2 \\
& I=\int \sqrt{(2-x)(3+x) d x}=\int \sqrt{6-x-x^2 d x} \\
& 6-x-x^2=-\left(x^2+x-6\right) \\
& =-\left\{x^2+2\left(\frac{1}{2}\right) x-6+\frac{1}{4}-\frac{1}{4}\right\} \\
& =-\left[\left(x+\frac{1}{2}\right)^2-\frac{25}{4}\right]=\frac{25}{4}-\left(x+\frac{1}{2}\right)^2 \\
& \therefore \quad \int \sqrt{6-x-x^2} d x=\int \sqrt{\frac{25}{4}-\left(x+\frac{1}{2}\right)^2} d x \\
& =\int \sqrt{\left(\frac{5}{2}\right)^2-\left(x+\frac{1}{2}\right)^2} d x \\
& =\frac{x+\frac{1}{2}}{2} \sqrt{6-x-x^2}+\frac{\frac{25}{4}}{2} \sin ^{-1}\left(\frac{x+\frac{1}{2}}{\frac{5}{2}}\right) \\
& =\frac{2 x+1}{4} \sqrt{6-x-x^2}+\frac{25}{8} \sin ^{-1}\left(\frac{2 x+1}{5}\right)+c \\
&
\end{aligned}$