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REPEATER 2024 — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsREPEATER 20244 Marks
Question
Evaluate: $\int x^2 \sin \ 3 x \ d x$
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Answer

Let $ I=\int x^2 \sin \ 3 x \ d x$
$=x^2 \int \sin 3 x \cdot\ d x-\int\left[\frac{d}{d x}\left(x^2\right) \int \sin 3 x \cdot\ d x\right] d x$
$\left[\because \int u v \cdot d x=u \int v \cdot d x-\int\left[\frac{d u}{d x} \int v \cdot\ d x\right] d x\right]$
$=x^2\left(-\frac{\cos 3 x}{3}\right)-\int 2 x\left(-\frac{\cos 3 x}{3}\right) d x$
$=-\frac{x^2}{3} \cos 3 x+\frac{2}{3} \int x \cos\ 3 x\ d x$
$=-\frac{x^2}{3} \cos 3 x+\frac{2}{3}\left[x \int \cos\ 3 x\ d x-\int\left\{\frac{d}{d x}(x) \int \cos \ 3 x \cdot d x\right\} d x\right]$
$\left[\because \int u v \cdot\ d x=u \int v \cdot\ d x-\int\left[\frac{d u}{d x} \int v \cdot\ d x\right] d x\right]$
$=-\frac{x^2}{3} \cos 3 x+\frac{2}{3}\left[\frac{x \sin 3 x}{3}-\int \frac{\sin 3 x}{3}\ d x\right]$
$=-\frac{x^2}{3} \cos 3 x+\frac{2 x \sin 3 x}{9}-\frac{2}{9} \int \frac{\sin 3 x}{3}\ d x$
$=-\frac{x^2}{3} \cos 3 x+\frac{2 x \sin 3 x}{9}-\frac{2}{9}\left(-\frac{\cos 3 x}{3}\right)+c$
$\int x^2 \sin\ 3 x\ d x$
$=-\frac{x^2}{3} \cos 3 x+\frac{2 x \sin 3 x}{9}+\frac{2 \cos 3 x}{27}+c$

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