# Math Help - Trigonometric substitution.

1. ## Trigonometric substitution.

hi, i have an example

$\int \sqrt{4^2-4^2\sin ^2\theta .4\cos \theta d\theta }$
it become
$\int 4^2 \cos ^2\theta d\theta$
which later becomes
$4^2\int (\cos 2\theta +1)/2.d\theta$

why is this so?

2. Originally Posted by salohcinseah
hi, i have an example

$\int \sqrt{4^2-4^2\sin ^2\theta .4\cos \theta d\theta }$
it become
$\int 4^2 \cos ^2\theta d\theta$
which later becomes
$4^2\int (\cos 2\theta +1)/2.d\theta$

why is this so?
That's only true if the original integral was $\int \sqrt{4^2-4^2\sin^2\theta}\cdot 4\cos\theta\,d\theta$, which is what I think you wanted to write in the first place.

First, we see that

\begin{aligned}\sqrt{4^2-4^2\sin^2\theta}&=\sqrt{4^2(1-\sin^2\theta)}\\&=\sqrt{4^2\cos^2\theta}\\&=4\cos \theta\end{aligned}

So the integral becomes $\int 4\cos\theta\cdot 4\cos\theta\,d\theta = 4^2\int\cos^2\theta\,d\theta$

And by the identity $\cos^2\theta=\tfrac{1}{2}(1+\cos(2\theta))$, we get $4^2\int\tfrac{1}{2}(1+\cos(2\theta))\,d\theta$, which is what we wanted.

Does this make sense?

3. yes, i understand now , the example skip a few step which make me lost