Find $\int \sin^4(x)\cos^5(x)dx$

Question

Find

$\int \sin^4(x)\cos^5(x)dx$

After $t=\tan(\frac{x}{2})$ I get

$32\int\frac{t^4(1-t^2)^5}{(1+t^2)^{10}}$

And I think to do this integral I need to do another trig substitution.Is there a different way of solving first integral?

Answer 1

Yes. Note that$$\sin^4(x)\cos^5(x)=\sin^4(x)\bigl(1-\sin^2(x)\bigr)^2\cos(x).$$So, do $\sin(x)=y$ and $\cos(x)\,\mathrm dx=\mathrm dy$.

Answer 2

A variant: linearise the integrand: \begin{align} \sin^4x\cos^5x&=\Bigl(\frac12\sin2x\Bigr)^{\mkern-4mu 4}\cos x=\frac1{16}\biggl(\frac{\mathrm e^{2ix}-\mathrm e^{-2ix}}{2i}\biggr)^{\mkern-6mu4} \frac{\mathrm e^{ix}-\mathrm e^{-ix}}{2} \\ &= \frac1{256}\frac{(\mathrm e^{8ix}-4\mathrm e^{4ix}+6-4\mathrm e^{-4ix}+\mathrm e^{-8ix})(\mathrm e^{ix}+\mathrm e^{-ix})}{2} \\ &=\dots=\frac1{256}\biggl[\begin{aligned}[t]\frac{\mathrm e^{9ix}+\mathrm e^{9ix}}2 &-4\,\frac{\mathrm e^{5ix}+\mathrm e^{-5ix}}2 + 6\,\frac{\mathrm e^{ix}+\mathrm e^{-ix}}2 \\ &-4\,\frac{\mathrm e^{3ix}+\mathrm e^{-3ix}}2 + \frac{\mathrm e^{7ix}+\mathrm e^{-7ix}}2\biggr] \end{aligned}\\ &= \frac1{256}\Bigl( \cos 9x +\cos 7x-4\cos 5x-4\cos 3x +6\cos x\Bigr) \end{align}

Answer 3

\begin{aligned} & \int \sin ^{4} x \cos ^{5} x d x \\ =& \int \sin ^{4} x \cos ^{4} x d(\sin x) \\ =& \int \sin ^{4} x\left(1-\sin ^{2} x\right)^{2} d(\sin x) \\ =& \int\left(\sin ^{4} x-2 \sin ^{6} x+\sin ^{8} x\right) d(\sin x) \\ =& \frac{\sin ^{5} x}{5}-\frac{2 \sin ^{7} x}{7}+\frac{\sin ^{9} x}{9}+C \end{aligned}