integration

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February 3rd 2007, 06:35 PM
viet

integration

need a little help with this problem:
$\int_{0}^{\pi/2} \dfrac{\sin\theta}{1+\cos^{2}\theta} d\theta$
February 3rd 2007, 07:14 PM
ThePerfectHacker

Quote:

Originally Posted by viet

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

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,
$\int \frac{\sin x}{1+\cos^2 x} dx$
Let $u=\cos x$ then $u'=-\sin x$
Substitution rule,
$- \int \frac{1}{1+u^2} du$
This, is the arctangent integral,
$-\tan^{-1}u+C=-\tan^{-1} (\cos x)+C=$
February 3rd 2007, 07:17 PM
AfterShock

Quote:

Originally Posted by viet

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

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

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