Exercise 11.8 - Chapter 11 Integral Calculus 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 11.8
Integrate the following with respect to $\mathrm{x}$.
Question 1.
(i) $e^{\mathrm{ax}} \cos b x$
(ii) $e^{2 x} \sin x$
(iii) $\mathrm{e}^{-\mathrm{x}} \cos 2 \mathrm{x}$
Solution:
(i) $\int e^{a x} \cos b x d x$
Let
$
\begin{aligned}
\mathrm{I} & =\int e^{a x} \cos b x d x \\
u & =\cos b x \\
d u & =-b \sin b x d x \quad \Rightarrow \quad d \nu=e^{a x} d x
\end{aligned}
$
Applying Integration by parts, .
$
\begin{aligned}
\mathrm{I} & =\frac{e^{a x}}{a} \cos b x-\int \frac{e^{a x}}{a}(-b \sin b x) d x \\
\mathrm{I} & =\frac{e^{a x}}{a} \cos b x+\frac{b}{a} \int e^{a x} \sin b x d x \\
u & =\sin b x \quad \Rightarrow d v=e^{a x} d x \\
d u & =b \cos b x d x \quad \Rightarrow \quad v=\frac{e^{a x}}{a}
\end{aligned}
$
Again Applying Integration by Parts,
$
\begin{aligned}
& \mathrm{I}=\frac{e^{a x}}{a} \cos b x+\frac{b}{a}\left[\frac{e^{a x}}{a} \sin b x-\frac{b}{a} \int e^{a x} \cos b x d x\right] \\
& \mathrm{I}=\frac{e^{a x} \cos b x}{a}+\frac{b}{a}\left[\frac{e^{a x} \sin b x}{a}-\frac{b}{a} \mathrm{I}\right] \\
& \mathrm{I}=\frac{e^{a x} \cos b x}{a}+\frac{b e^{a x} \sin b x}{a^2}-\frac{b^2}{a^2} \mathrm{I}
\end{aligned}
$

$
\begin{aligned}
\mathrm{I}+\frac{b^2}{a^2} \mathrm{I} & =\frac{a e^{a x} \cos b x+b e^{a x} \sin b x}{a^2}+c \\
\mathrm{I}\left(\frac{a^2+b^2}{a^2}\right) & =\frac{e^{a x}[a \cos b x+b \sin b x]}{a^2}+c \\
\mathrm{I} & =\frac{e^{a x}}{a^2+b^2}[a \cos b x+b \sin b x]+c \\
\therefore \quad \int e^{a x} \cos b x d x & =\frac{e^{a x}}{a^2+b^2}[a \cos b x+b \sin b x]+c
\end{aligned}
$
(ii) $\int e^{2 x} \sin x d x$
Using the formula
$
\int e^{2 x} \sin x d x=\frac{e^{a x}}{a^2+b^2}[a \sin b x-b \cos b x]+c
$
for $a=2, b=1$, we get
$
\begin{aligned}
\int e^{2 x} \sin x d x & =\frac{e^{2 x}}{2^2+1^2}[2 \sin x-\cos x]+c \\
& =\frac{e^{2 x}}{5}[2 \sin x-\cos x]+c
\end{aligned}
$
(iii) $\int e^{-x} \cos 2 x$
Using the formula
$
\int e^{a x} \cos b x d x=\frac{e^{a x}}{a^2+b^2}[a \cos b x+b \sin b x]+c
$
for $a=-1, b=2$, we get
$
\begin{aligned}
\int e^{-x} \cos 2 x d x & =\frac{e^{-x}}{(-1)^2+(2)^2}[-\cos 2 x+2 \sin 2 x]+c \\
& =\frac{e^{-x}}{5}[-\cos 2 x+2 \sin 2 x]+c
\end{aligned}
$
Question 2.
(i) $e^{-3 x} \sin 2 x$
(ii) $e^{-4 x} \sin 2 x$
(iii) $e^{-3 x} \cos x$
Solution:

(i) $\int e^{-3 x} \sin 2 x d x$
Using the formula
$
\text { . } \int e^{a x} \sin b x d x=\frac{e^{a x}}{a^2+b^2}[a \sin b x-b \cos b x]+c
$
for $a=-3, b=2$, we get
$
\begin{aligned}
\int e^{-3 x} \sin 2 x d x & =\frac{e^{-3 x}}{(-3)^2+(2)^2}[-3 \sin 2 x-2 \cos 2 x]+c \\
& =\frac{-e^{-3 x}}{13}[3 \sin 2 x+2 \cos 2 x]+c
\end{aligned}
$
(ii) $\int e^{-4 x} \sin 2 x d x$
Using the formula
$
\int e^{a x} \sin b x d x=\frac{e^{a x}}{a^2+b^2}[a \sin b x-b \cos 2 x]+c
$
for $a=-4, b=2$, we get
$
\begin{aligned}
\int e^{-4 x} \sin 2 x d x & =\frac{-e^{-4 x}}{20}[4 \sin 2 x+2 \cos 2 x]+c \\
& =\frac{-e^{-4 x}}{10}[2 \sin 2 x+\cos 2 x]+c
\end{aligned}
$
(iii) $\int e^{-3 x} \cos x d x$
Using the formula
$
\int e^{a x} \cos b x d x=\frac{e^{a x}}{a^2+b^2}[a \cos b x+b \sin b x]+c
$
for $a=-3, b=1$, we get
$
\int e^{-3 x} \cos x d x=\frac{e^{-3 x}}{10}[-3 \cos x+\sin x]+c
$