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Naveen Sharma 8 years, 1 month ago
{tex}\int \sin ^3 x\cos ^3 x dx\\ = \int \sin ^3 x(1- \sin^2x)\cos x dx\\ put \sin x = t\ then\ \cos x dx = dt {/tex}
{tex}= \int t^3 (1- t^2) dt\\ = \int (t^3- t^5) dt\\ = \int t^3 dt - \int t^5 dt\\ = {t^4\over 4} - {t^6\over 6}+ C \\ = {\sin ^4x\over 4} - {\sin^6x\over 6}+ C {/tex}
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