integrate

is 4/15 the ans??

given is as
\int_{0}^{\frac{\pi }{2}}{sin^{4}x.sinx}dx
=\int_{0}^{\frac{\pi }{2}}{(1-cos^{2}x)^{2}.sinx}dx
=\int_{0}^{\frac{\pi }{2}}{(1-2cos^{2}x+cos^{4}x).sinx}dx
=\int_{0}^{\frac{\pi }{2}}{sinx}dx +\int_{0}^{\frac{\pi }{2}}{cos^{4}x.sinx}dx - 2\int_{0}^{\frac{\pi }{2}}{cos^{2}x.sinx}dx
=1 -\left[\frac{cos^{5}x}{5} \right]^{\pi /2}_{0}+\left[\frac{cos^{3}x}{3} \right]^{\pi /2}_{0}
=1 +\frac{1}{5}-\frac{2}{3} = \frac{8}{15}

what is the gamma function????

The gamma function is the following definite integral -:
\int_{0}^{\pi /2}{sin^mx cos^nx}dx = \frac{[(m-1)(m-3)....2,1][(n-1)(n-3)...2,1]}{[(m+n)(m+n-2)...2,1]}

2,1 means the series ends as 2 or 1 as the situation demands.
The important rule here is that we multiply the above result by pi/2 ONLY and ONLY IF both m and n are even integers.

So your integral comes out as
\int_{0}^{\pi /2}{sin^5x}dx = \frac{4.2}{5.3.1} = \frac{8}{15}
As the cosine term is not present, we've not used the "n" series and put n = 0 in the denominator.

Good one Pritish..here again ur starting another revolution..[3]

okay got it....this is really nice...gonna save a loooot of time

A small info-- gamma functions are also known as "Walli's formula"[1]

∫_o^{Π/2}2sin^2m-1xcos^2n-1xdx=Γ(m)Γ(n)/Γ(m+n)

Γ(m)=(m-1)Γ(m-1)

Γ(1/2)=√Î