Exercise 11.9 - Chapter 11 - Integral Calculus - 11th Maths Guide Samacheer Kalvi Solutions

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Ex 11.9
Integrate the following with respect to $\mathrm{x}$
Question 1.
$\mathrm{e}^{\mathrm{x}}(\tan \mathrm{x}+\log \sec \mathrm{x})$
Solution:
Let
Take
$
\begin{aligned}
\mathrm{I} & =\int e^x(\tan x+\log \sec x) d x \\
f(x) & =\log \sec x \\
f^{\prime}(x) & =\frac{1}{\sec x} \times \sec x \tan x=\tan x
\end{aligned}
$
This is of the form $\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c$
$
\therefore \quad \int e^x(\log \sec x+\tan x) d x=e^x \log |\sec x|+c .
$
Question 2.
$
\mathrm{e}^{\mathrm{x}}\left(\frac{x-1}{2 x^2}\right)
$
Solution:
$
\text { Let } \begin{aligned}
I & =\frac{1}{2} \int e^x\left(\frac{x}{x^2}-\frac{1}{x^2}\right) d x \\
& =\frac{1}{2} \int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x
\end{aligned}
$
Take
$
f(x)=\frac{1}{x} \text { and } f^{\prime}(x)=-\frac{1}{x^2}
$
This is of the form $\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c$
$
\begin{aligned}
& =\frac{1}{2} e^x\left[\frac{1}{x}\right]+c \\
& =\frac{e^x}{2 x}+c
\end{aligned}
$
Question 3.
$
\mathrm{e}^{\mathrm{x}} \sec \mathrm{x}(1+\tan \mathrm{x})
$
Solution:
Let $\mathrm{I}=\mathrm{I}=\int e^x(\sec x+\sec x \tan x) d x$

Take $f(x)=\sec \mathrm{x}$
$f^{\prime}(x)=\sec x \tan x$
This is of the form of $\int e^x\left[f(x)+f^{\prime}(x)\right] d x=\mathrm{e}^{\mathrm{x}} \mathrm{f}(\mathrm{x})+\mathrm{c}$ $=\mathrm{e}^{\mathrm{x}} \sec \mathrm{x}+\mathrm{c}$
Question 4.
$\mathrm{e}^{\mathrm{x}}\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right)$
Solution:
Let
$
\begin{aligned}
& I=\int e^x\left(\frac{2+2 \sin x \cos x}{1+\cos 2 x}\right) d x \\
& =\int 2 e^x \frac{(1+\sin x \cos x)}{2 \cos ^2 x} d x \\
& =\int e^x\left[\frac{1}{\cos ^2 x}+\frac{\sin x \cos x}{\cos ^2 x}\right] d x \\
& =\int e^x\left[\sec ^2 x+\tan x\right] d x
\end{aligned}
$
Take
$
\begin{aligned}
f(x) & =\tan x \\
f^{\prime}(x) & =\sec ^2 x
\end{aligned}
$
This is of the form of $\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c$
$
=e^x \tan x+c
$
Question 5.
$
e^{\tan ^{-1} x}\left(\frac{1+x+x^2}{1+x^2}\right)
$

Solution:
$
\begin{aligned}
& \int e^{\tan ^{-1} x}\left(\frac{1+x+x^2}{1+x^2}\right) d x \\
& \text { put } \\
& t=\tan ^{-1} x \\
& \frac{d t}{d x}=\frac{1}{1+x^2} \\
& \left(1+x^2\right) d t=d x \\
&
\end{aligned}
$
Substituting (2) and (3) in (1)
$
\begin{aligned}
& \Rightarrow \quad \int e^t\left(\frac{1+x+x^2}{1+x^2}\right)\left(1+x^2\right) d t=\int e^t\left(1+\tan t+\tan ^2 t\right) d t \quad \begin{array}{l}
t=\tan ^{-1} x \\
\tan t=x
\end{array} \\
& =\int e^t\left[\tan t+\sec ^2 t\right] d t \quad\left(1+\tan ^2 t=\sec ^2 t\right) \\
&
\end{aligned}
$
This is of the form of $\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c$
$
=e^t \tan x
$
$
\begin{aligned}
\text { (Replace } t & \left.=\tan ^{-1} x .\right) \\
& =e^{\tan ^{-1} x} \tan \left(\tan ^{-1} x\right) \\
& =x e^{\tan ^{-1} x}+c .
\end{aligned}
$

Question 6.
$
\frac{\log x}{(1+\log x)^2}
$
Solution:
$
\begin{aligned}
& I=\int \frac{\log x}{(1+\log x)^2} d x \\
& \text { put } t=\log x ; \quad e^t=x ; \quad e^t d t=d x \text {. } \\
& \mathrm{I}=\int \frac{t}{(1+t)^2} e^t d t=\int \frac{t+1-1}{(1+t)^2} e^t d t \\
& =\int e^t\left[\frac{t+1}{(1+t)^2}-\frac{1}{(1+t)^2}\right] d t=\int e^t\left[\frac{1}{1+t}-\frac{1}{(1+t)^2}\right] d t \\
&
\end{aligned}
$
Let
This is of the form of $\int e^t\left[f(t)+f^{\prime}(t)\right] d t=e^t f(t)+c$
$
\begin{aligned}
& =e^t\left[\frac{1}{1+t}\right]+c \\
& =x\left[\frac{1}{1+\log |x|}\right]+c \\
& =\frac{x}{1+\log |x|}+c .
\end{aligned}
$