$\int \sin x (1 + \cos^{2} x) dx$

By Tony Ho (P1942) on Tuesday, September 5, 2000 - 12:15 am :

How do you integrate:

$\int \sin x (1 + \cos^{2} x) dx$ ?? I have been trying to do this for quite a while now, but I must say I am really stuck! I guess you have to make some substitutions using the product formulae or the sum formulae or the trigonometric identities or things like that, but I cannot do it. I appreciate if someone can help.

Tony Ho

By Pras Pathmanathan (Pp233) on Tuesday, September 5, 2000 - 01:37 am :

You don't have to worry about substituting an identity for the

\cos^{2} x

. Instead just try the substitution

u = \cos x

So $du / dx = - \sin x$ , so $- du = \sin x dx$ , and we have that:

$\int \sin x (1 + \cos^{2} x) dx$

$= - \int 1 + u^{2} du$

$= - u - u^{3} / 3 +$ constant

$= - \cos x - (\cos^{3} x) / 3) +$ constant

This works because you'll notice that we have ( $- 1$ times) the derivative of $\cos x$ times a function of $\cos x$ , so when we do the substitution, the $\sin x$ disappears nicely.

Hope that helps,

Pras

By Tony Ho (P1942) on Tuesday, September 5, 2000 - 03:09 pm :

Thanks very much, I am ashamed I haven't noticed something so obvious.