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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]4 Marks
Question
$\int x \sin 2 x \cos 5 x d x$
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Answer

$\text { Let } I =\int x \sin 2 x \cos 5 x d x$
$=\frac{1}{2} \int x(2 \sin 2 x \cos 5 x) d x$
$=\frac{1}{2} \int x[\sin (2 x+5 x)+\sin (2 x-5 x)] d x$
$=\frac{1}{2} \int x[\sin 7 x-\sin (-3 x)] d x$
$=\frac{1}{2} \int x(\sin 7 x-\sin 3 x) d x$
$=\frac{1}{2} \int x \sin 7 x d x-\frac{1}{2} \int x \sin 3 x d x$
$=\frac{1}{2}\left[x \int \sin 7 x d x-\int\left\{\frac{ d }{ d x}(x) \int \sin 7 x d x\right\} d x\right]-\frac{1}{2}\left[x \int \sin 3 x d x-\int\left\{\frac{ d }{ d x}(x) \int \sin 3 x d x\right\} d x\right]$
$=\frac{1}{2}\left[x\left(-\frac{\cos 7 x}{7}\right)-\int 1 \cdot\left(\frac{-\cos 7 x}{7}\right) d x\right]-\frac{1}{2}\left[x\left(\frac{-\cos 3 x}{3}\right)-\int 1 \cdot\left(\frac{-\cos 3 x}{3}\right) d x\right]$
$=\frac{1}{2}\left(\frac{-x \cos 7 x}{7}+\frac{1}{7} \int \cos 7 x d x\right)-\frac{1}{2}\left(\frac{-x \cos 3 x}{3}+\frac{1}{3} \int \cos 3 x d x\right)$
$=\frac{1}{2}\left[\frac{-x \cos 7 x}{7}+\frac{1}{7}\left(\frac{\sin 7 x}{7}\right)\right]-\frac{1}{2}\left[\frac{-x \cos 3 x}{3}+\frac{1}{3}\left(\frac{\sin 3 x}{3}\right)\right]+ c $

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