# Math Help - Integration by Parts

1. ## Integration by Parts

Hi How do I do integrate:

(sinx)^n * (cosx)^2 when it is a definite integral between x = pi/2 and x = 0

2. Originally Posted by Sunyata
Hi How do I do integrate:

(sinx)^n * (cosx)^2 when it is a definite integral between x = pi/2 and x = 0
Put cos(x) = t
-sin(x)dx = dt
dx= - dt/sin(x)
So Integration I = - intg[(sinx)^n*t^2*dt/sin(x)
= - intg[(sinx)^(n-1)*t^2*dt.
Now do the integration by parts.

3. This is NOT a candidate for "integration by parts". I would be inclined to use the fact that $cos^2(x)= 1- sin^2(x)$ and write
$\int_0^{\pi/2} sin^n(x)(1- sin^2(x))dx= \int_0^{\pi/2} sin^n(x)dx- \int_0^{\pi/2} sin^{n+2}(x) dx$.

Those can be done by the "reduction integrals" given in any Calculus book.

4. hi
$\int_{0}^{\frac{\pi }{2}}(\sin^n(x)\cos^2(x))dx$= $\int_{0}^{\frac{\pi }{2}}(\sin^n(x)\cos(x))(\cos(x))dx$= $\left [ \frac{\sin^{n+1}}{n+1}\cos(x) \right ]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\frac {\sin^{n+2}(x)}{n+1}dx$= $\int_{0}^{\frac{\pi}{2}}\frac{\sin^{n+2}(x)}{n+1}d x=\frac{1}{n+1}\int_{0}^{\frac{\pi}{2}}\sin^{n+2}( x)dx$