1. ## Trigonometric Integral

Sorry, I'm new and I don't know how to show mathematical symbols on here.

The integral from 0 to pi/2 of sin^4x dx. The exponent is just 4, not 4x.

Thanks a lot!

2. Originally Posted by element
Sorry, I'm new and I don't know how to show mathematical symbols on here.

The integral from 0 to pi/2 of sin^4x dx. The exponent is just 4, not 4x.

Thanks a lot!
$\int_0^{\frac{\pi}{2}} \sin^4 x\, dx = \int_0^{\frac{\pi}{2}} \sin^2 x(1 - \cos^2x)\, dx$

you can use the following identities:

$\cos 2x = \cos^2 x - \sin^2x = 2\cos^2x - 1 = 1 - 2\sin^2x$

3. Originally Posted by kalagota
$\int_0^{\frac{\pi}{2}} \sin^4 x\, dx = \int_0^{\frac{\pi}{2}} \sin^2 x(1 - \cos^2x)\, dx$

you can use the following identities:

$\cos 2x = \cos^2 x - \sin^2x = 2\cos^2x - 1 = 1 - 2\sin^2x$
Uhm...why rewrite it as that?

$\int\sin^4(x)dx=\int\left(\frac{1-\cos(2x)}{2}\right)^2dx$

Expand that and see that

$\cos^2(nx)=\frac{1+\cos(2nx)}{2}$

4. Originally Posted by Mathstud28
Uhm...why rewrite it as that?

$\int\sin^4(x)dx=\int\left(\frac{1-\cos(2x)}{2}\right)dx$

Expand that and see that

$\cos^2(nx)=\frac{1+\cos(2nx)}{2}$
How is
$sin^4(x) = \frac{1 - cos(2x)}{2}$?

-Dan