Integrals — Class 12 Maths Solution

: Let $I = \int {{{\sin }^3}\left( {2x + 1} \right)} dx$

$= \cfrac{1}{4}\int {\left[ {3\sin \left( {2x + 1} \right) - \sin 3\left( {2x + 1} \right)} \right]dx}$

$= \cfrac{3}{4}\left( { - \cfrac{{\cos \left( {2x + 1} \right)}}{2}} \right) - \cfrac{1}{4}\left( {\cfrac{{ - \cos 3\left( {2x + 1} \right)}}{6}} \right) + C$

$= - \cfrac{3}{8}\cos \left( {2x + 1} \right) + \cfrac{1}{{24}}\cos 3\left( {2x + 1} \right) + C$

$= - \cfrac{3}{8}\cos \left( {2x + 1} \right) + \cfrac{1}{{24}}\left[ {4{{\cos }^3}\left( {2x + 1} \right) - 3\cos \left( {2x + 1} \right)} \right] + C$

$= - \cfrac{3}{8}\cos \left( {2x + 1} \right) + \cfrac{1}{6}{\cos ^3}\left( {2x + 1} \right) - \cfrac{1}{8}\cos \left( {2x + 1} \right) + C$

$= - \cfrac{1}{2}\cos \left( {2x + 1} \right) + \cfrac{1}{6}{\cos ^3}\left( {2x + 1} \right) + C$