Exercise 11.8-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.
$
\mathrm{e}^{2 \mathrm{x}} \sin 3 \mathrm{xdx}
$
Solution:

Let
(i.e.,)
$
\begin{aligned}
\mathrm{I} & =\int e^{2 x} \sin 3 x d x \quad \text { put } \quad u=\sin 3 x ; \quad d u=3 \cos 3 x d x \\
e^{2 x} d x & =d \nu \\
& =(\sin 3 x)\left(\frac{e^{2 x}}{2}\right)-\int \frac{e^{2 x}}{2}(3 \cos 3 x) d x \\
& =\frac{e^{2 x}}{2} \sin 3 x-\frac{3}{2} \int e^{2 x} \cos 3 x d x=\frac{e^{2 x}}{2} \sin 3 x-\frac{3}{2} \mathrm{I}_1 \\
\mathrm{I} & =\int e^{2 x} \sin 3 x d x \\
& =\frac{e^{a x}}{a^2+b^2}(a \sin b x-b \cos b x)=\frac{e^{a x}}{13}[2 \sin 3 x-3 \cos 3 x]+c . \\
\mathrm{I}_1 & =\int e^{2 x} \cos x 3 x d x \\
u & =\cos 3 x ; \quad d u=-3 \sin 3 x d x \\
e^{2 x} d x & =d v, \quad \therefore \quad v=\frac{e^{2 x}}{2} \\
\mathrm{I}_1 & =(\cos 3 x) \frac{e^{2 x}}{2}-\int \frac{e^{2 x}}{2}(-3 \sin 3 x) d x \\
\mathrm{I}_1 & =\frac{e^{2 x}}{2} \cos 3 x+\frac{3}{2} \int e^{2 x} \sin 3 x d x \\
& =\frac{e^{2 x}}{2} \cos 3 x+\frac{3}{2} \mathrm{I}
\end{aligned}
$

Substituting (2) in (1) we get,
$
I=\frac{e^{2 x}}{2} \sin 3 x-\frac{3}{2}\left\{\frac{e^{2 x}}{2} \cos 3 x+\frac{3}{2} \mathrm{I}\right\}
$
(i.e.,)
$
I=\frac{e^{2 x}}{2}\left[\sin 3 x-\frac{3}{2} \cos 3 x\right]-\frac{9}{4} I
$
(i.e.,)
$
\begin{aligned}
I\left(1+\frac{9}{4}\right) & =\frac{e^{2 x}}{2}[2 \sin 3 x-3 \cos 3 x] \\
I\left(\frac{13}{4}\right) & =\frac{e^{2 x}}{2}[2 \sin 3 x-3 \cos 3 x]
\end{aligned}
$
So,
(i.e.,)
$
\begin{aligned}
& \mathrm{I}=\frac{4}{13} \frac{e^{2 x}}{4}[2 \sin 3 x-3 \cos 3 x] \\
& \mathrm{I}=\frac{e^{2 x}}{13}[2 \sin 3 x-3 \cos 3 x]+c .
\end{aligned}
$

Question 2.
$\mathrm{e}^{\mathrm{x}} \cos 2 \mathrm{x}$
Solution:
$
I=\int e^x \cos 2 x d x
$
Let
$
\begin{aligned}
u & =\cos 2 x ; \quad d u & =-2 \sin 2 x d x \\
e^x d x & =d v \quad \therefore \quad v & =e^x
\end{aligned}
$
Now
$
\begin{aligned}
I & =(\cos 2 x) e^x-\int e^x(-2 \sin 2 x) d x \\
& =e^x \cos 2 x+2 \int e^x \sin 2 x d x
\end{aligned}
$
(i.e.,) $\mathrm{I}=e^x \cos 2 x+2 \mathrm{I}_1$
Formouatabethod:
$
I=\int e^x \cos 2 x d x=\frac{e^x}{5}(\cos 2 x+2 \sin 2 x)+c
$
Formula:
$
\begin{aligned}
& =\frac{e^{a x}}{a^2+b^2}[a \cos b x+b \sin b x] \\
\mathrm{I}_1 & =\int e^x \sin 2 x d x
\end{aligned}
$
Let
$
u=\sin 2 x ; \quad d u=2 \cos 2 x d x
$
$
e^x d x=d \nu \quad \therefore \quad v=e^x
$
$
\text { Now } \quad \begin{aligned}
\mathrm{I}_1 & =\int e^x \sin 2 x d x=\int \sin 2 x d\left(e^x\right) \\
& =(\sin 2 x) e^x-\int e^x(2 \cos 2 x) d x \\
& =e^x \sin 2 x-2 \mathrm{I}_1
\end{aligned}
$
Now
Substituting (2) in (1) we get,
$
\begin{aligned}
& \mathrm{I}=\mathrm{e}^{\mathrm{x}} \cos 2 \mathrm{x}+\left[\mathrm{e}^{\mathrm{x}}+\sin 2 \mathrm{x}-2 \mathrm{I}\right] \\
& \mathrm{I}=\mathrm{e}^{\mathrm{x}} \cos 2 \mathrm{x}+2 \mathrm{e}^{\mathrm{x}} \sin 2 \mathrm{x}-4 \mathrm{I} \\
& 5 \mathrm{I}=\mathrm{e}^{\mathrm{x}}(\cos 2 \mathrm{x}+2 \sin 2 \mathrm{x}) \\
& \therefore \mathrm{I}=\frac{\mathrm{e}^x}{5}[\cos 2 \mathrm{x}+2 \sin 2 \mathrm{x}]+\mathrm{c}
\end{aligned}
$
Question 3.
$\mathrm{e}^{3 \mathrm{x}} \sin 2 \mathrm{x}$
Solution:

$
I=\int e^{3 x} \sin 2 x d x
$
Put
(i.e.,)
$
\begin{aligned}
u=\sin 2 x ; & d u=2 \cos 2 x d x \\
e^{3 x} d x=d \nu \quad & \therefore \quad \quad \quad v=\frac{e^{3 x}}{3} \\
\mathrm{I} & =(\sin 2 x)\left(\frac{e^{3 x}}{3}\right)-\int \frac{e^{3 x}}{3} 2 \cos 2 x d x \\
\mathrm{I} & =\frac{e^{3 x}}{3} \sin 2 x-\frac{2}{3} \mathrm{I}_1
\end{aligned}
$
Formula method:
$
\begin{aligned}
\mathrm{I} & =\int e^{a x} \cdot \sin b x \cdot d x=\frac{e^{a x}}{a^2+b^2}[a \sin b x-b \cos b x]+c \\
\mathrm{I} & =\int e^{3 x} \cdot \sin 2 x d x=\frac{e^{3 x}}{3}[3 \sin 2 x-2 \cos 2 x]+c \\
\mathrm{I}_1 & =\int e^{3 x} \cos 2 x d x \\
u & =\cos 2 x ; d u=-2 \sin 2 x d x \\
e^{3 x} d x= & \therefore \quad \quad \quad v=\frac{e^{3 x}}{3} \\
\therefore \quad \mathrm{I}_1 & =\int(\cos 2 x)\left(\frac{e^{3 x}}{3}\right)-\int \frac{e^{3 x}}{3}(-2 \sin 2 x) d x \\
& =\frac{e^{3 x}}{3} \cos 2 x+\frac{2}{3} \mathrm{I}
\end{aligned}
$
Substituting (2) in (1) we get,
$
\begin{aligned}
& \mathrm{I}=\frac{e^{3 x}}{3} \sin 2 x-\frac{2}{3}\left\{\frac{e^{3 x}}{3} \cos 2 x+\frac{2}{3} \mathrm{I}\right\} \\
& \mathrm{I}=\frac{e^{3 x}}{9}\{3 \sin 2 x-2 \cos 2 x\}-\frac{4}{9} \mathrm{I}
\end{aligned}
$

$\begin{equation}
\begin{aligned}
I\left(1+\frac{4}{9}\right) & =\frac{e^{3 x}}{9}[3 \sin 2 x-2 \cos 2 x] \\
I\left(\frac{13}{9}\right) & =\frac{e^{3 x}}{9}[3 \sin 2 x-2 \cos 2 x] \\
\therefore \quad I & =\frac{9}{13} \times \frac{e^{3 x}}{9}[3 \sin 2 x-2 \cos 2 x] \\
& =\frac{e^{3 x}}{13}[3 \sin 2 x-2 \cos 2 x]+c .
\end{aligned}
\end{equation}$