Exercise 11.6 - Chapter 11 Integral Calculus 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 11.6
Integrate the following with respect to $\mathrm{x}$.
Question 1.
$
\int \frac{x}{\frac{x}{\sqrt{1+x^2}}} d x \text { put } 1+x^2=t
$
D.w.r. to ' $x$ '
$
\begin{array}{r}
0+2 x=\frac{d t}{d x} \\
d x=\frac{d t}{2 x}
\end{array}
$
sub (2) and (3) in (1)
$
\begin{aligned}
\Rightarrow(1) \Rightarrow \quad \int \frac{x}{\sqrt{t}} \frac{d t}{2 x} & =\frac{1}{2} \int \frac{1}{\sqrt{t}} d t \\
& =\frac{1}{2} \int \frac{1}{t^{1 / 2}} d t=\frac{1}{2}\left[\frac{-1}{-\frac{1}{t^{-1 / 2}}}\right]+c \\
& =\sqrt{t}+c=\sqrt{1+x^2}+c
\end{aligned}
$
Question 2.
$
\frac{x^2}{1+x^6}
$

Solution:
$
\begin{aligned}
& \int \frac{x^2}{1+x^6} d x=\int \frac{x^2}{1+\left(x^3\right)^2} d x \\
& \qquad t=x^3
\end{aligned}
$
D.w.r. to ' $x$ '
$
\begin{aligned}
\frac{d t}{d x} & =3 x^2 \\
\frac{d t}{3 x^2} & =d x
\end{aligned}
$
sub (2) and (3) in (1)
$
\begin{aligned}
\Rightarrow \quad \int \frac{\chi^2}{1+t^2} \frac{d t}{3 \chi^2} & =\frac{1}{3} \int \frac{d t}{1+t^2}=\frac{1}{3} \tan ^{-1}(t)+c \\
& =\frac{1}{3} \tan ^{-1}\left(x^3\right)+c
\end{aligned}
$
Question 3.
$\frac{e^x-e^{-x}}{e^x+e^{-x}}$
Solution:
$
\int \frac{e^x-e^{-x}}{e^x+e^{-x}} d x \quad t=e^x+e^{-x}
$
D.w.r. to ' $x$ '
$
\begin{aligned}
\frac{d t}{d x} & =e^x-e^{-x} \\
\frac{d t}{e^x-e^{-x}} & =d x
\end{aligned}
$
sub (2) and (3) in (1)
$
\begin{aligned}
& \Rightarrow \int \frac{\left(e^x-e^-x\right)}{t} \times \frac{d t}{\left(e^x-e^-x\right)}=\int \frac{1}{t} d t=\log t+c \\
& =\log \left(e^x+e^{-x}\right)+c \\
&
\end{aligned}
$

Question 4.
$
\frac{10 x^9+10^x \log _e 10}{10^x+x^{10}}
$

Solution:
$
\int \frac{10 x^9+10^x \log _e 10}{10^x+x^{10}} d x
$
$
t=10^x+x^{10}
$
D.w.r. to ' $x$ '
$
\begin{aligned}
\frac{d t}{d x} & =10^x \log _e 10+10 x^9 \\
\frac{d t}{10^x \log _e 10+10 x^9} & =d x
\end{aligned}
$
sub (2) and (3) in (1)
$
\begin{aligned}
\Rightarrow \int \frac{\left(10 x^9+10^x \log _e 10\right)}{t} \times \frac{d t}{\left(10^x \log _e 10+10 x^9\right)} & =\int \frac{1}{t} d t=\log t+c \\
& =\log \left|10^x+x^{10}\right|+c
\end{aligned}
$

Question 5.
$
\frac{\sin \sqrt{x}}{\sqrt{x}}
$
Solution:
$
\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x
$
$
\text { D.w.r. to ' } x \text { ' } \begin{aligned}
t & =\sqrt{x} \\
\frac{d t}{d x} & =\frac{1}{2 \sqrt{x}} \\
2 \sqrt{x} d t & =d x
\end{aligned}
$
,sub (2) and (3) in (1)
$
\Rightarrow \int \frac{\sin t}{\sqrt{x}} \times 2 \sqrt{x} d t=2 \int \sin t d t=2[-\cos t]+c=-2 \cos \sqrt{x}+c
$
Question 6.
$
\frac{\cot x}{\log (\sin x)}
$

Solution:
$
\begin{aligned}
& \int \frac{\cot x}{\log (\sin x)} d x . \\
& \text { D.w.r. to ' } x \text { ' } \quad t=\log (\sin x)
\end{aligned}
$
$
\begin{aligned}
\frac{d t}{d x} & =\frac{1}{\sin x} \times \cos x \\
\frac{d t}{d x} & =\cot x \\
\frac{d t}{\cot x} & =d x
\end{aligned}
$
sub (2) and (3) in (1)
$
\begin{aligned}
\Rightarrow \quad \int \frac{\cot x}{t} \cdot \frac{d t}{\cot x} & =\int \frac{1}{t} d t \\
& =\log t+c=\log |\log (\sin x)|+c
\end{aligned}
$
Question 7.
$
\frac{\operatorname{cosec} x}{\log \left(\tan \frac{x}{2}\right)}
$

Solution:
$
\begin{aligned}
& \int \frac{\operatorname{cosec} x}{\log \left(\tan \frac{x}{2}\right)} d x \\
& \text { D.w.r. to ' } x \text { ' } t=\log \left(\tan \frac{x}{2}\right)
\end{aligned}
$
$
\begin{aligned}
\frac{d t}{d x} & =\frac{1}{\tan \frac{x}{2}} \times \sec ^2 \frac{x}{2} \times \frac{1}{2} \\
\frac{d t}{d x} & =\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \times \frac{1}{\cos ^2 \frac{x}{2}} \times \frac{1}{2} \\
\frac{d t}{d x} & =\frac{1}{\sin x} \\
\frac{d t}{d x} & =\operatorname{cosec} x \\
\frac{d t}{\operatorname{cosec} x} & =d x
\end{aligned}
$
sub (2) and (3) in (1)
$
\Rightarrow \int \frac{\operatorname{cosec} x}{t} \cdot \frac{d t}{\operatorname{cosec} x}=\int \frac{1}{t} d t=\log t+c=\log \left|\log \left(\tan \frac{x}{2}\right)\right|+c .
$
Question 8 .
$
\frac{\sin 2 x}{a^2+b^2 \sin ^2 x}
$
Solution:

$
\int \frac{\sin 2 x}{a^2+b^2 \sin ^2 x} d x \quad t=a^2+b^2 \sin ^2 x
$
D.w.r. to ' $x$ '
$
\begin{aligned}
\frac{d t}{d x} & =0+b^2(2 \sin x \cos x) \\
\frac{d t}{d x} & =b^2 \sin 2 x \\
\frac{d t}{b^2 \sin 2 x} & =d x
\end{aligned}
$
sub (2) and (3) in (1)
$
\begin{aligned}
\Rightarrow \quad \int \frac{\sin 2 x}{t} \cdot \frac{d t}{b^2 \sin 2 x} & =\frac{1}{b^2} \int \frac{1}{t} d t \\
& \left.=\frac{1}{b^2} \log t+c=\frac{1}{b^2} \log \mid a^2+b^2 \sin ^2 x\right) \mid+c .
\end{aligned}
$
Question 9.
$
\frac{\sin ^{-1} x}{\sqrt{1-x^2}}
$
Solution:
$
\begin{aligned}
& \int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x \\
& \frac{d t}{d x}=\frac{1}{\sqrt{1-x^2}} \\
& \sqrt{1-x^2} d t=d x
\end{aligned}
$
sub (2) and (3) in (1)
$
\begin{aligned}
\Rightarrow \int \frac{t}{\sqrt{1-x^2}} \sqrt{1-x^2} d t & =\int t d t \\
& =\frac{t^2}{2}+c=\frac{\left(\sin ^{-1} x\right)^2}{2}+c .
\end{aligned}
$

Question 10.
$\frac{\sqrt{x}}{1+\sqrt{x}}$
Solution:
$
\int \frac{\sqrt{x}}{1+\sqrt{x}} d x \quad \begin{aligned}
t & =1+\sqrt{x} \\
\frac{d t}{d x} & =0+\frac{1}{2 \sqrt{x}} \\
2 \sqrt{x} d t & =d x
\end{aligned}
$
sub (2) and (3) in (1)
$
\begin{aligned}
\Rightarrow \quad \int \frac{\sqrt{x}}{t} 2 \sqrt{x} d t & =2 \int \frac{x}{t} d t \\
& =2 \int \frac{(t-1)^2}{t} d t \\
& =2 \int \frac{\left(t^2-2 t+1\right)}{t} d t \\
& =2 \int\left[t-2+\frac{1}{t}\right] d t=2\left[\frac{t^2}{2}-2 t+\log t\right]+c \\
& =2\left[\frac{(1+\sqrt{x})^2}{2}-2(1+\sqrt{x})+\log |1+\sqrt{x}|\right]+c \\
& =(1+\sqrt{x})^2-4(1+\sqrt{x})+2 \log |1+\sqrt{x}|+c
\end{aligned}
$
Question 11.
$\frac{1}{x \log x \log (\log x)}$

Solution:
$
\begin{aligned}
& \int \frac{1}{x \log x \log (\log x)} d x \\
& t=\log (\log x) \\
& \text { D.w.r. to ' } x \text { ' } \\
& \frac{d t}{d x}=\frac{1}{\log x} \times \frac{1}{x} \\
& x \log x d t=d x \\
& \text { sub (2) and (3) in (1) } \\
& \Rightarrow \quad \int \frac{1}{x \log x t} x \log x d t=\int \frac{1}{t} d t \\
& =\log t+c=\log |\log (\log x)|+c \text {. } \\
&
\end{aligned}
$
Question 12.
$\alpha \beta \mathrm{x}^{\alpha-1} \mathrm{e}^{-\beta \mathrm{x}^\alpha}$
Solution:
$\int \alpha \beta x^{\alpha-1} e^{-\beta x^\alpha} d x$
$
t=\beta x^\alpha
$
D.w.r. to ' $x$ '
$
\begin{aligned}
\frac{d t}{d x} & =\alpha \beta x^{\alpha-1} \\
\frac{d t}{\alpha \beta x^{\alpha-1}} & =d x
\end{aligned}
$
sub (2) and (3) in (1)
$
\begin{aligned}
\Rightarrow \quad \int \alpha \beta x^{\alpha-1} e^{-t} \frac{d t}{\alpha \beta x^{\alpha-1}} & =\int e^{-t} d t \\
& =-e^{-t} d t+c=-e^{-\beta x^\alpha}+c .
\end{aligned}
$
Question 13.
$\tan x \sqrt{\sec x}$
Solution:

$
\begin{aligned}
& \int \tan x \sqrt{\sec x} d x \\
& t=\sec x \\
& \frac{d t}{d x}=\sec x \tan x \\
& \frac{d t}{\sec x \tan x}=d x \\
&
\end{aligned}
$
sub (2) and (3) in (1)
$
\begin{aligned}
\Rightarrow \quad \int \tan x \sqrt{t} \frac{d t}{\sec x \tan x} & =\int \frac{\sqrt{t}}{t} d t=\int \frac{1}{\sqrt{t}} d t \\
& =2 \sqrt{t}+c=2 \sqrt{\sec x}+c .
\end{aligned}
$
Question 14.
$x(1-x)^{17}$
Solution:
$
\begin{aligned}
& \int x(1-x)^{17} d x \\
& \Rightarrow \quad t=1-x \\
& 1-t=x
\end{aligned}
$
D.w.r. to ' $x$ '
$
\begin{gathered}
\frac{d t}{d x}=-1 \\
-d t=d x
\end{gathered}
$
sub (2) and (3) in (1)
$
\begin{aligned}
\Rightarrow \quad \int(1-t) t^{17}(-d t) & =-\int\left(t^{17}-t^{18}\right) d t=\int\left(t^{18}-t^{17}\right) d t \\
& =\frac{t^{19}}{19}-\frac{t^{18}}{18}+c=\frac{(1-x)^{19}}{19}-\frac{(1-x)^{18}}{18}+c .
\end{aligned}
$
Question 15.
$\sin ^5 x \cos ^3 x$

Solution:
$
\begin{aligned}
\int \sin ^5 x \cos ^3 x d x=\int \sin ^5 x \cos x & \left(\cos ^2 x\right) d x \\
& =\int \sin ^5 x \cos x\left(1-\sin ^2 x\right) d x \\
& \left.=\int \sin ^5 x \cos x-\sin ^7 x \cos x\right) d x \\
\int\left[f(x)^n\right] f^{\prime}(x) d x & =\frac{[f(x)]^{n+1}}{n+1} \\
& =\frac{(\sin x)^6}{6}-\frac{(\sin x)^8}{8}+c=\frac{\sin ^6 x}{6}-\frac{\sin ^8 x}{8}+c
\end{aligned}
$
Question 16.
$
\begin{aligned}
& \frac{\cos x}{\cos (x-a)} \\
& \text { Solution: } \\
& \int \frac{\cos (x-a+a)}{\cos (x-a)} d x=\int\left[\frac{\cos (x-a) \cos a}{\cos (x-a)}-\frac{\sin (x-a) \sin a}{\cos (x-a)}\right] d x \\
& =\int[\cos a-\tan (x-a) \sin a] d x \\
& =x \cos a-\log |\sec (x-a)| \times \sin a+c \\
& =x \cos a-\sin a \log |\sec (x-a)|+c \text {. } \\
&
\end{aligned}
$