# Math Help - Integration with trig functions

1. ## Integration with trig functions

1) Integral of the inverse of sinx from 0 to 1/2
2) Integral of (sinx)^3 * (cosx)^2

1) Is there an identity I should be looking for when dealing with inverse sine?

2) Should I be using integration by parts in this?

2. 1) Integration by parts

2) $\int\sin^3x\cos^2xdx=\int(1-\cos^2x)\cos^2x\sin xdx$
Put $\cos x=u$

3. Originally Posted by red_dog
1) Integration by parts

2) $\int\sin^3x\cos^2xdx=\int(1-\cos^2x)\cos^2x\sin xdx$
Put $\cos x=u$
should it be just (1-cos^2x) or that over 2? In my textbook it reads:

sin^2x = (1-cos^2x) / 2

4. $\sin^2x+\cos^2x=1\Rightarrow\sin^2x=1-\cos^2x$
$\displaystyle\sin^2x=\frac{1-\cos 2x}{2}$

5. Originally Posted by red_dog
$\sin^2x+\cos^2x=1\Rightarrow\sin^2x=1-\cos^2x$
$\displaystyle\sin^2x=\frac{1-\cos 2x}{2}$
thanks.
for the first one, what do I set dV equal to if i have u = sin x ?

6. $\displaystyle\int_0^{\frac{1}{2}}\arcsin xdx=\int_0^{\frac{1}{2}}x'\arcsin xdx=\left.x\arcsin x\right|_0^{\frac{1}{2}}-\int_0^{\frac{1}{2}}\frac{x}{\sqrt{1-x^2}}dx=$
$\displaystyle=\frac{\pi}{12}+\left.\sqrt{1-x^2}\right|_0^{\frac{1}{2}}=\frac{\pi}{12}+\frac{\ sqrt{3}}{2}-1$

7. Originally Posted by xfyz
1) Integral of the inverse of sinx from 0 to 1/2
$\int\arcsin x\,dx$ (integration limits omitted).

Apply integration by parts, so

$u=\arcsin x\implies du=\frac1{\sqrt{1-x^2}}\,dx$ & $dv=dx\implies v=x$

I assume you know integration by parts formula, this yields

$\int\arcsin x\,dx=x\arcsin x-\int\frac x{\sqrt{1-x^2}}\,dx$

For the remaining integral, let's set $y=\sqrt{1-x^2}\implies dy=-\frac x{\sqrt{1-x^2}}\,dx$

Therefore $\int\arcsin x\,dx=x\arcsin x+\sqrt{1-x^2}+k$

Then we happily can say that

$\int_0^{1/2}\arcsin x\,dx=\left.x\arcsin x+\sqrt{1-x^2}\right|_0^{1/2}$, yielding red_dog's answer