Exercise 11.5-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
Integrate the following

Question 1 .
$
\frac{e^{2 x}+e^{-2 x}+2}{e^x}
$

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

Question 4.
$\cos 2 \mathrm{x} \sin 4 \mathrm{x}$
Solution:

$
\begin{aligned}
\sin 4 x \cos 2 x & =\frac{1}{2}[2 \sin 4 x \cos 2 x]=\frac{1}{2}\{\sin (4 x+2 x)+\sin (4 x-2 x)\} \\
= & \frac{1}{2}[\sin 6 x+\sin 2 x] \\
\therefore \int \cos 2 x \sin 4 x d x & =\frac{1}{2} \int(\sin 6 x+\sin 2 x) d x \\
& =\frac{1}{2}\left\{\frac{-\cos 6 x}{6}+\frac{(-\cos 2 x)}{2}\right\}=\frac{-1}{2}\left[\frac{\cos 6 x}{6}+\frac{\cos 2 x}{2}\right]+c .
\end{aligned}
$
Question 5.
$\left(e^x-1\right)^2 e^{-4 x}$ Solution:
$
\begin{aligned}
\left(e^{2 x}+1-2 e^x\right) e^{-4 x} & =e^{-2 x}+e^{-4 x}-2 e^{-3 x} \\
\int\left(e^x-1\right)^2 e^{-4 x} d x & =\int\left(e^{-2 x}+e^{-4 x}-2 e^{-3 x}\right) d x \\
& =\frac{e^{-2 x}}{-2}+\frac{e^{-4 x}}{-4}-2 \frac{e^{-3 x}}{-3}=-\left\{\frac{e^{-2 x}}{2}+\frac{e^{-4 x}}{4}-2 \frac{e^{-3 x}}{3}\right\} \\
& =-\left\{\frac{e^{-2 x}}{2}+\frac{e^{-4 x}}{4}-2 \frac{e^{-3 x}}{3}\right\}+c .
\end{aligned}
$
Question 6
$
(\mathrm{x}+1) \sqrt{x+3}
$
Solution:
$
\begin{aligned}
& \int(x+1) \sqrt{x+3} d x \\
& \text { put } x+3=t ; \quad d x=d t \\
& x=t-3 ; \quad x+1=t-3+1=t-2 \\
& \text { Now } \quad \mathrm{I}=\int(x+1) \sqrt{x+3} d x=\int(t-2) \sqrt{t} d t \\
&=\int\left(t^{3 / 2}-2 t^{1 / 2}\right) d t=\frac{t^{5 / 2}}{5 / 2}-\frac{2 t^{3 / 2}}{3 / 2} \\
&=\frac{2}{5}(x+3)^{5 / 2}-\frac{4}{3}(x+3)^{3 / 2}+c .
\end{aligned}
$

Question 7.
$
(2 \mathrm{x}+1) \sqrt{2 x+3}
$
Solution:

$\begin{aligned} & \int(2 x+1) \sqrt{2 x+3} d x \\ & \text { put } 2 x+3=t ; \quad 2 d x=d t \\ & \therefore \quad d x=\frac{d t}{2} \\ & 2 x+3=t \quad \therefore \quad 2 x=t-3 \\ & 2 x+1=t-3+1=t-2 \\ & \text { Now, } \quad \mathrm{I}=\int(t-2) \sqrt{t}\left(\frac{d t}{2}\right) \\ & =\frac{1}{2} \int\left(t^{3 / 2}-2 t^{1 / 2}\right) d t=\frac{1}{2}\left\{\frac{t^{5 / 2}}{5 / 2}-\frac{2 t^{3 / 2}}{3 / 2}\right\} \\ & =\frac{1}{2}\left\{\frac{2}{5} t^{5 / 2}-\frac{4}{3} t^{3 / 2}\right\}=\frac{1}{5}(2 x+3)^{5 / 2}-\frac{2}{3}(2 x+3)^{3 / 2}+c \\ & \end{aligned}$