Show $\int^{\pi/2}_0 \cos^{\mu}(x)\sin^{v}(x)dx= \frac{1}{2}B(\frac{1+\mu}{2},\frac{1+v}{2})$

Question

I have the following task:

The Beta integral $B(p,q)= \int^1_0t^{p-1}(1-t)^{q-1}dt$ is convergent for all $p,q \gt 0$

Check that $\int^{\pi/2}_0\cos^{\mu}(x)\sin^{v}(x)dx= \frac{1}{2}B(\frac{1+\mu}{2},\frac{1+v}{2})$ for $\mu,v \gt -1$

What I tried so far:

First of all I calculated the RHS of the equality:

$$\frac{1}{2}B(\frac{1+\mu}{2},\frac{1+v}{2})= \frac{1}{2}\int^1_0t^{\frac{1+\mu}{2}-1}(1-t)^{\frac{1+v}{2}-1}dt=\frac{1}{2}\int^1_pt^{\frac{\mu-1}{2}}(1-t)^{\frac{v-1}{2}}dt=\frac{1}{2}\int^1_0t^{\frac{\mu-1}{2}}dt-\frac{1}{2}\int^1_0t^{\frac{\mu+v-2}{2}}dt=\frac{1}{2}\frac{t^{\frac{\mu-1}{2}+1}}{\frac{\mu-1}{2}+1}\Big|^1_0-\frac{1}{2}\frac{t^{\frac{\mu+v}{2}}}{\frac{\mu+v}{2}}\Big|^1_0=\frac{t^{\frac{\mu+1}{2}}}{\mu+1}\Big|^1_0-\frac{t^{\mu+v}}{\frac{\mu+v}{2}}\Big|^1_0=\frac{1}{\mu+1}-\frac{1}{\mu+v}=\frac{\mu+v-\mu-1}{(\mu+v)(\mu+1)}=\frac{v-1}{(\mu+v)(\mu+1)}$$

But when I try to calculate $\int^{\pi/2}_0\cos^{\mu}(x)\sin^{v}(x)dx$, whatever I try, I can't seem to get the result. I've looked on Wikipedia for the integral, and the value of it is:

\begin{aligned} \int\left(\sin ^{n} a x\right)\left(\cos ^{m} a x\right) d x &=-\frac{\left(\sin ^{n-1} a x\right)\left(\cos ^{m+1} a x\right)}{a(n+m)}+\frac{n-1}{n+m} \int\left(\sin ^{n-2} a x\right)\left(\cos ^{m} a x\right) d x \quad \,(\text { for } m, n>0) \\ &=\frac{\left(\sin ^{n+1} a x\right)\left(\cos ^{m-1} a x\right)}{a(n+m)}+\frac{m-1}{n+m} \int\left(\sin ^{n} a x\right)\left(\cos ^{m-2} a x\right) d x \quad \,(\text { for } m, n>0) \end{aligned}

And I don't think this is even close to what I'm trying to get. What am I doing wrong? Any help is appreciated!

Answer 1

Taking the substitution $ \small\left\lbrace\begin{aligned}t&=\cos^{2}{x}\\ \mathrm{d}t&=-2\sin{x}\cos{x}\,\mathrm{d}t\end{aligned}\right. $, we get : $$ \small\int_{0}^{\frac{\pi}{2}}{\cos^{\mu}{x}\sin^{\nu}{x}\,\mathrm{d}x}=\frac{1}{2}\int_{\frac{\pi}{2}}^{0}{\left(\cos^{2}{x}\right)^{\frac{1+\mu}{2}-1}\left(1-\cos^{2}{x}\right)^{\frac{1+\nu}{2}-1}\left(-2\sin{x}\cos{x}\right)\mathrm{d}x}=\frac{1}{2}\int_{0}^{1}{t^{\frac{1+\mu}{2}-1}\left(1-t\right)^{\frac{1+\nu}{2}-1}\,\mathrm{d}t} $$

Answer 2

You went wrong when you wrote the integral as the difference of two other integrals; you appear to have mistaken $(1-t)^{(\nu-1)/2}$ with $1-t^{(\nu-1)/2}$. As @Gary noted, $t=\sin^2x$ finishes the job, viz. $dt=2\sin x\cos xdx$.