trig integration

[ijd5000]

i have an integral with an even power of sine and cosine i'm looking for the general pattern how to solve these, i know it has to involve the half angel identities.

Heres a specific problem:

The function is: 3sin^2x cos^4 x dx from pi to 0.

 

[daon2]

The standard method is to replace using the half angle identities:

\(\displaystyle \sin^2(x)=\dfrac{1-\cos(2x)}{2}\)

\(\displaystyle \cos^2(x)=\dfrac{1+\cos(2x)}{2}\)

so

\(\displaystyle \sin^2x \cos^4x = \left(\dfrac{1-\cos(2x)}{2}\right)\left(\dfrac{1+\cos(2x)}{2} \right)^2\) which is ugly, but doable. Repeated applications of the identity will be needed.

Alternatively:

\(\displaystyle \sin^2x \cos^4x = \sin^2x(1-\sin^2x)^2 = \sin^2x-2\sin^4x+\sin^6x\)

It might also be useful in some cases to use \(\displaystyle \sin x\cos x = \frac{1}{2}\sin(2x)\) for integrals such as

\(\displaystyle \displaystyle \int \sin^4x\cos^4x dx = \frac{1}{16} \int \sin^4(2x) dx\)

 

[Deleted member 4993]

Anotherway:

sin²(x) * cos⁴(x) = cos⁴(x) - cos⁶(x)

cos⁴(x) = 3/8 + 1/2 * cos(2x) + 1/8 * cos(4x)

cos⁶(x) = 5/16 + 1/32 * [cos(6x) + 6 * cos(4x) + 15 * cos(2x)]

and so on.....