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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals3 Marks
Question
Find the integrals of the function sin 3x cos 4x
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Answer

$\int {\sin 3x\cos 4xdx = \frac{1}{2}\int {2\sin 3x\cos 4xdx} }$

$ = \frac{1}{2}\int {\left\{ {\sin \left( {4x + 3x} \right) - \sin \left( {4x - 3x} \right)} \right\}dx} $  [Using 2 sin B cos A = sin (A + B) - sin (A - B)]

$= \frac{1}{2}\int {\left( {\sin 7x - \sin x} \right)dx} $

$ = \frac{1}{2}\left[ {\int {\sin 7xdx - \int {\sin xdx} } } \right]$

$= \frac{1}{2}\left[ {\frac{{ - \cos 7x}}{7} - \left( { - \cos x} \right)} \right] + c$

$= \frac{{ - 1}}{{14}}\cos 7x + \frac{1}{2}\cos x + c$

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