Reduction formula Integration

Question

Is it true that $I_{p,q}=\left ( \frac{p-1}{p+q} \right )I_{p-q,q}$ given that $I_{p,q}=\int_{0}^{\frac{\pi}{2}}\sin^{p}x \cos^q x dx$ (where $p$ and $q$ are positive integers)? If so, how? Also, how can I evaluate $\int_{0}^{\frac{\pi}{2}}\sin^{p}x \cos^q x dx$ ($p$ and $q$ are positive integers) using the very expressions from the above question?

Answer 1

You need not use a reduction formula. Here's why. Consider the function defined by $$\Gamma(s)=\int_0^\infty t^{s-1}e^{-t}dt\qquad s>0.$$ Integration by parts shows that $$\Gamma(s+1)=s\Gamma(s),$$ and it is easily shown that $\Gamma(1)=1$. Thus for $s=1,2,3,...$, $$\Gamma(s)=(s-1)!.$$ This function is known as the Gamma function. Anyway, we set $t=x^2$ and see that $$\Gamma(s)=2\int_0^\infty x^{2s-1}e^{-x^2}dx.$$ So we see $$\Gamma(a)\Gamma(b)=4\int_0^\infty \int_0^\infty y^{2a-1}x^{2b-1}e^{-(x^2+y^2)}dydx.$$ Then we convert to polar coordinates and get $$\begin{align} \Gamma(a)\Gamma(b)&=4\int_0^{\pi/2}\int_0^\infty \sin(\theta)^{2a-1}\cos(\theta)^{2b-1}e^{-r^2}r^{2(a+b)-1}drd\theta\\ &=2\int_0^\infty e^{-r^2}r^{2(a+b)-1}2\int_0^{\pi/2}\sin(\theta)^{2a-1}\cos(\theta)^{2b-1}d\theta dr\\ &=\left(2\int_0^\infty e^{-r^2}r^{2(a+b)-1}dr\right)\left(2\int_0^{\pi/2}\sin(\theta)^{2a-1}\cos(\theta)^{2b-1}d\theta\right)\\ &=2\Gamma(a+b)\int_0^{\pi/2}\sin(\theta)^{2a-1}\cos(\theta)^{2b-1}d\theta. \end{align}$$ So what we have ended up showing is that $$\int_0^{\pi/2}\sin(t)^{a}\cos(t)^{b}dt=\frac{\Gamma(\tfrac{a+1}2)\Gamma(\tfrac{b+1}2)}{2\Gamma(\tfrac{a+b}2+1)}.$$