Integration with trig functions

**xfyz** · September 3rd 2007, 08:55 AM

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?

**red_dog** · September 3rd 2007, 09:01 AM

1) Integration by parts

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

**xfyz** · September 3rd 2007, 11:58 AM

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

**red_dog** · September 3rd 2007, 12:07 PM

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

**xfyz** · September 3rd 2007, 12:10 PM

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 ?

**red_dog** · September 3rd 2007, 12:17 PM

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

**Krizalid** · September 3rd 2007, 01:25 PM

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

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