need a little help with this problem:

$\displaystyle \int_{0}^{\pi/2} \dfrac{\sin\theta}{1+\cos^{2}\theta} d\theta$

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- Feb 3rd 2007, 06:35 PMvietintegration
need a little help with this problem:

$\displaystyle \int_{0}^{\pi/2} \dfrac{\sin\theta}{1+\cos^{2}\theta} d\theta$ - Feb 3rd 2007, 07:14 PMThePerfectHacker
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,

$\displaystyle -\tan^{-1}u+C=-\tan^{-1} (\cos x)+C=$ - Feb 3rd 2007, 07:17 PMAfterShock
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