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Indefinite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integration4 Marks
Question
Evaluate : $\int e^{2 x} \cdot \sin 3 x \cdot d x$
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Answer

$\quad \mathrm{I}=\int e^{2 x \cdot} \sin 3 x \cdot d x$
Here we use repeated integration by parts.
To evaluate $\int e^{a x} \cdot \sin (b x+c) \cdot d x ; \int e^{a x} \cdot \cos (b x+c) \cdot d x$ any function can be taken as a first function.
$
\begin{aligned}
& \mathrm{I}=e^{2 x} \cdot \int \sin 3 x \cdot d x-\int \frac{d}{d x} \cdot e^{2 x} \cdot \int \sin 3 x \cdot d x \cdot d x \\
& =e^{2 x} \cdot\left(-\cos 3 x \cdot \frac{1}{3}\right)-\int e^{2 x \cdot 2}\left(-\cos 3 x \cdot \frac{1}{3}\right) \cdot d x \\
& =-\frac{1}{3} \cdot e^{2 x} \cdot \cos 3 x+\frac{2}{3} \int e^{2 x} \cdot \cos 3 x \cdot d x \\
& =-\frac{1}{3} \cdot e^{2 x} \cdot \cos 3 x+\frac{2}{3}\left(e^{2 x} \cdot \int \cos 3 x \cdot d x-\int \frac{d}{d x} \cdot e^{2 x} \cdot \int \cos 3 x \cdot d x \cdot d x\right) \\
& =-\frac{1}{3} \cdot e^{2 x} \cdot \cos 3 x+\frac{2}{3}\left[e^{2 x \cdot}\left(\sin 3 x \cdot \frac{1}{3}\right)-\int e^{2 x \cdot 2} \cdot\left(\sin 3 x \cdot \frac{1}{3}\right) \cdot d x\right] \\
& =-\frac{1}{3} \cdot e^{2 x} \cdot \cos 3 x+\frac{2}{9} \cdot e^{2 x} \cdot \sin 3 x-\frac{4}{9} \cdot \int e^{2 x} \cdot \sin 3 x \cdot d x \\
& I=-\frac{1}{3} \cdot e^{2 x} \cdot \cos 3 x+\frac{2}{9} \cdot e^{2 x} \cdot \sin 3 x-\frac{4}{9} \cdot I \\
\end{aligned}
$
$\begin{aligned} & \mathrm{I}+\frac{4}{9} \cdot \mathrm{I}=\frac{e^{2 x}}{9}[-3 \cos 3 x+2 \sin 3 x]+c \\ & \frac{13}{9} \cdot \mathrm{I}=\frac{e^{2 x}}{9}[2 \sin 3 x-3 \cos 3 x]+c\end{aligned}$
$\begin{gathered}=\frac{e^{2 x}}{13}[2 \sin 3 x-3 \cos 3 x]+c \\ \therefore \int e^{2 x \cdot} \sin 3 x \cdot d x=\frac{e^{2 x}}{13}[2 \sin 3 x-3 \cos 3 x]+c\end{gathered}$

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