calc II: Let I_n = int [0 to pi/2] [sin^n(x)] dx, and show

[PaulErly]

I got this question in math and I have no idea where to really begin. One thing that I's sorta basic, yet still perplexing, is what exactly does I sub n mean? Specifically, what is the difference between I sub n and I sub 2n.

Thanks!
Paul

 

[galactus]

Perhaps start with:

\(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}{sin^{n}(x)}dx=\frac{n-1}{n}\int_{0}^{\frac{\pi}{2}}{sin^{n-2}(x)}dx\)

 

[pka]

This problem really has very little to do with integrals apart from this.
For \(\displaystyle \L
x \in [a,b],\quad f(x) \le g(x) \Rightarrow \int\limits_a^b f \le \int\limits_a^b g .\)

Here is the trick: for each
\(\displaystyle \L
x \in \left[ {0,\frac{\pi }{2}} \right],\quad 0 \le \sin (x) \le 1 \Rightarrow 0 \le \sin ^{n + 1} (x) \le \sin ^n (x).\)

Now all one has to do is put the two together.