hi, i have an example

$\displaystyle \int \sqrt{4^2-4^2\sin ^2\theta .4\cos \theta d\theta } $

it become

$\displaystyle \int 4^2 \cos ^2\theta d\theta $

which later becomes

$\displaystyle 4^2\int (\cos 2\theta +1)/2.d\theta $

why is this so?

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- May 21st 2011, 12:25 AMsalohcinseahTrigonometric substitution.
hi, i have an example

$\displaystyle \int \sqrt{4^2-4^2\sin ^2\theta .4\cos \theta d\theta } $

it become

$\displaystyle \int 4^2 \cos ^2\theta d\theta $

which later becomes

$\displaystyle 4^2\int (\cos 2\theta +1)/2.d\theta $

why is this so? - May 21st 2011, 12:39 AMChris L T521
That's only true if the original integral was $\displaystyle \int \sqrt{4^2-4^2\sin^2\theta}\cdot 4\cos\theta\,d\theta$, which is what I think you wanted to write in the first place.

First, we see that

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

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

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

Does this make sense? - May 21st 2011, 12:54 AMsalohcinseah
yes, i understand now , the example skip a few step which make me lost