Exercise 11.6-Additional Questions - 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.
$
(2 \mathrm{x}+3) \sqrt{x^2+3 x-5}
$
Solution:
$
\begin{aligned}
& \int(2 x+3) \sqrt{x^2+3 x-5} d x \\
& \text { put } t=x^2+3 x-5 ; \quad d t=(2 x+3) d x \\
& \mathrm{I}=\int \sqrt{t} d t=\int t^{1 / 2} d t=\frac{t^{3 / 2}}{3 / 2}=\frac{2}{3} t^{3 / 2}=\frac{2}{3}\left(x^2+3 x-5\right)^{3 / 2}+c
\end{aligned}
$
Question 2.
$\frac{x \sin ^{-1}\left(x^2\right)}{\sqrt{1-x^4}}$

Solution:
$
\begin{aligned}
& \int \frac{x \sin ^{-1}\left(x^2\right)}{\sqrt{1-x^4}} d x \\
& \text { put } \sin ^{-1}\left(x^2\right)=t \\
& \frac{1}{\sqrt{1-x^4}}(2 x d x)=d t \quad \therefore \quad \frac{x d x}{\sqrt{1-x^4}}=\frac{d t}{2} \\
& \mathrm{I}=\int t\left(\frac{d t}{2}\right)=\frac{1}{2} \int t d t=\frac{1}{2}\left(\frac{t^2}{2}\right)=\frac{1}{4} t^2=\frac{1}{4}\left(\sin ^{-1} x^2\right)^2+c
\end{aligned}
$
Question 3.
$\sec 4 \mathrm{x} \tan \mathrm{x}$
Solution:
$
\begin{aligned}
\int_0 \sec ^4 x \tan x d x & =\int \sec ^3 x(\sec x \tan x) d x \\
\text { put } t & =\sec x ; \quad d t=\sec x \tan x d x \\
I & =\int t^3 d t=\frac{t^4}{4}=\frac{(\sec x)^4}{4}=\frac{1}{4} \sec ^4 x+c
\end{aligned}
$
Question 4.
$\frac{\sin x}{\sin (x+a)}$
Solution:
$
\begin{aligned}
\int \frac{\sin x}{\sin (x+a)} d x & =\int \frac{\sin (x+a-a)}{\sin (x+a)} d x \\
& =\int \frac{\sin (x+a) \cos a-\cos (x+a) \sin a}{\sin (x+a)} d x \\
& =\int \cos a d x-\int \sin a \cot (x+a) d x \\
& =(\cos a)(x)-\sin a[\log \sin (x+a)] \\
& =x \cos a-\sin a \log \sin (x+a)+c
\end{aligned}
$

Question 5.
$\frac{\sqrt{\tan x}}{\sin x \cos x}$
Solution:

$\begin{aligned}
& \int \frac{\sqrt{\tan x}}{\sin x \cos x} d x=\int \frac{\sqrt{\tan x}}{\sin x \cos x} \cdot \frac{\sqrt{\tan x}}{\sqrt{\tan x}} d x \\
& =\int \frac{\tan x d x}{\sqrt{\tan x} \sin x \cos x}=\int \frac{d x}{\sqrt{\tan x} \cos ^2 x}=\int \frac{\sec ^2 x d x}{\sqrt{\tan x}} \\
& \text { put } t=\tan x ; \quad d t=\sec ^2 x d x \\
& =\int \frac{d t}{\sqrt{t}}=\int t^{-1 / 2} d t=\frac{t^{1 / 2}}{1 / 2}=2 \sqrt{t} \\
& =2 \sqrt{\tan x}+c \text {. } \\
&
\end{aligned}$

Question 6 .
$
x(1-x)^{16}
$
Solution:
$
\begin{aligned}
& \int x(l-x)^{16} d x \\
& \text { put } t=l-x \Rightarrow x=l-t ; \quad d t=-d x \quad \therefore \quad d x=-d t \\
& \mathrm{I}=\int(l-t) t^{16}(-d t)=\int-l t^{16}+t^{17} d t=\int\left(t^{17}-l t^{16}\right) d t \\
&=\frac{t^{18}}{18}-\frac{l t^{17}}{17}=\frac{(l-x)^{18}}{18}-\frac{l(l-x)^{17}}{17}+c
\end{aligned}
$

Question 7.
$
x^2(2-x)^{15}
$
Solution:
$
\begin{aligned}
\int x^2(2-x)^{15} d x & \\
\text { put } t & =2-x ;(\therefore \quad x=2-t) \\
d t & =-d x \text { So } d x=-d t \\
& \left.=\int(2-t)^2\left(t^{15}\right)\right)(-d t)=-\int\left(4+t^2-4 t\right) t^{15}(d t) \\
& =-\int\left(4 t^{15}+t^{17}-4 t^{16}\right) d t=\int\left(-t^{17}+4 t^{16}-4 t^{15}\right) d t \\
& =-\frac{t^{18}}{18}+\frac{4 t^{17}}{17}-\frac{4 t^{16}}{16}=\frac{-t^{18}}{18}+\frac{4}{17} t^{17}-\frac{1}{4} t^{16} \\
& =-\frac{1}{18}(2-x)^{18}+\frac{4}{17}(2-x)^{17}-\frac{1}{4}(2-x)^{16}+c
\end{aligned}
$
Question 8.
$
(x+1) \sqrt{2 x+3}
$

Solution:
$\int(x+1) \sqrt{2 x+3} d x$
put $t=2 x+3 ; \quad d t=2 d x \quad$ or $\quad d x=\frac{d t}{2}$
Now
$
\begin{gathered}
t=2 x+3 \Rightarrow \frac{t-3}{2}=x \\
\therefore \quad x+1=\frac{t-3}{2}+1=\frac{t-3+2}{2}=\frac{t-1}{2} \\
\mathrm{I}=\int\left(\frac{t-1}{2}\right)(\sqrt{t})\left(\frac{d t}{2}\right) \\
=\frac{1}{4} \int\left(t^{3 / 2}-t^{1 / 2}\right) d t=\frac{1}{4}\left\{\frac{t^{3 / 2}}{5 / 2}-\frac{t^{3 / 2}}{3 / 2}\right\} \\
=\frac{1}{4}\left\{\frac{2}{5} t^{5 / 2}-\frac{2}{3} t^{3 / 2}\right\}=\frac{1}{2}\left\{\frac{1}{5} t^{5 / 2}-\frac{1}{3} t^{3 / 2}\right\} \\
=\frac{1}{2}\left\{\frac{(2 x+3)^{5 / 2}}{5}-\frac{(2 x+3)^{3 / 2}}{3}\right\}+c
\end{gathered}
$
Question 9.
$\left(\mathrm{x}^2\right) \sqrt{x+1}$
Solution:
$
\begin{aligned}
\int\left(x^2+1\right) \sqrt{x+1} d x & \\
\text { put } t & =x+1 \Rightarrow x=t-1 ; \quad d t=d x \\
x^2+1 & =(t-1)^2+1=t^2-2 t+1+1=t^2-2 t+2 \\
\mathrm{I} & =\int\left(x^2+1\right) \sqrt{x+1} d x=\int\left(t^2-2 t+2\right) \sqrt{t} d t=\int\left(t^{5 / 2}-2 t^{3 / 2}+2 t^{1 / 2}\right) d t \\
& =\frac{t^{7 / 2}}{7 / 2}-\frac{2 t^{5 / 2}}{5 / 2}+\frac{2 t^{3 / 2}}{3 / 2}=\frac{2}{7} t^{7 / 2}-\frac{4}{5} t^{5 / 2}+\frac{4}{3} t^{3 / 2}=2\left\{\frac{1}{7} t^{1 / 2}-\frac{2}{5} t^{5 / 2}+\frac{2}{3} t^{3 / 2}\right\} \\
& =2\left\{\frac{1}{7}(x+1)^{7 / 2}-\frac{2}{5}(x+1)^{5 / 2}+\frac{2}{3}(x+1)^{3 / 2}\right\}+c
\end{aligned}
$