# Math Help - confusion with trig integrals

1. ## confusion with trig integrals

I seem to be coming across the same problem when I encounter an integral that requires using angle formulas to simplify the factors...such as

Indefinite integral: sin^2(2x)dx; I know the identity is [(1-cos(4x))/x]^2 which brings me to---> Indefinite Integral: 1/4 [1-2cos4x+cos^2(4x)] .... i get confused with the 1 b/c my last problem was The indefinite integral of sin^4(3t)dt where the result was 1/4 [1-2cos6t+1/2(cos12t)] dt and my answer book said it becomes 1/4 [3/2-2cos6t+1/2(cos12t)] --- I don't understand how they got 3/2 from 1 .... please help. thank you

2. Originally Posted by bgonzal8
I seem to be coming across the same problem when I encounter an integral that requires using angle formulas to simplify the factors...such as

Indefinite integral: sin^2(2x)dx; I know the identity is [(1-cos(4x))/x]^2 which brings me to---> Indefinite Integral: 1/4 [1-2cos4x+cos^2(4x)] .... i get confused with the 1 b/c my last problem was The indefinite integral of sin^4(3t)dt where the result was 1/4 [1-2cos6t+1/2(cos12t)] dt and my answer book said it becomes 1/4 [3/2-2cos6t+1/2(cos12t)] --- I don't understand how they got 3/2 from 1 .... please help. thank you
${\color{red}\boxed{{\color{black}{{\sin }^2}\theta = \frac{1}{2}\left( {1 - \cos 2\theta } \right)}}}$

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