1. ## integration

need a little help with this problem:
$\displaystyle \int_{0}^{\pi/2} \dfrac{\sin\theta}{1+\cos^{2}\theta} d\theta$

2. Originally Posted by viet
need a little help with this problem:
$\displaystyle \int_{0}^{\pi/2} \dfrac{\sin\theta}{1+\cos^{2}\theta} d\theta$
The evaluation at the limits is simple once you find the anti-derivative. Thus, I will find the anti-derivative and let you do the evaluation.

We have,
$\displaystyle \int \frac{\sin x}{1+\cos^2 x} dx$
Let $\displaystyle u=\cos x$ then $\displaystyle u'=-\sin x$
Substitution rule,
$\displaystyle - \int \frac{1}{1+u^2} du$
This, is the arctangent integral,
$\displaystyle -\tan^{-1}u+C=-\tan^{-1} (\cos x)+C=$

3. Originally Posted by viet
need a little help with this problem:
$\displaystyle \int_{0}^{\pi/2} \dfrac{\sin\theta}{1+\cos^{2}\theta} d\theta$
This will help:

int(1/(1 + x^2)) = arctan(x)

Let u = cos(x)

Then,

du = -sin(x)

-du = sin(x)

Factor out the -1

-int(1/(1 + u)^2)du

-arctan(u)du

Thus, substitute in for u:

-arctan(cos(x)) + C, but you're given the limits. And thus,

-arctan(cos(Pi/2)) - [-arctan(cos(0))]

0 - (-Pi/4)

= Pi/4