Evaluate the integrals \int sin^5xdx

\( \int sin^5x\;dx\)

The given equation can be written as

\( \int sin^5x\;dx= \int sin^3xsin^2x\;dx\)

\( \int (sin^3x-sin^3xcos^2x)dx\)

\( \int sin^xdx-\int sinx\;cos^2xdx-\int sin^3xcos^2xdx\)

We know that , d(cosx) = -sinx dx

So put cosx =t  and dt = -sinx dx  in above integrals

\( \int sin^xdx-\int sinx\;cos^2xdx-\int sin^3xcos^2xdx\)

\( \int sin^xdx+\int t^2dt+\int (t^2-t^4)dt\)

= \( -cosx+\dfrac{2}{3}cos^3x-\dfrac{1}{5}cos^5x+c\)