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Mathematics 9 Online
OpenStudy (anonymous):

Solving the flowing integral sin^3xcos2xdx

OpenStudy (anonymous):

\[\int\limits_{?}^{?} \sin^3xcos^2x dx\]

OpenStudy (anonymous):

Use the fact that sin^2=1-cos^2 to get sin^2xcos^x (1-cos^2x) and substitute u=cos x du=-sinx to get -integral u^2(1-u^2)du. you should be able to handle it from here

OpenStudy (anonymous):

sin^2xcos^2x sorry

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