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,
$\displaystyle -\tan^{-1}u+C=-\tan^{-1} (\cos x)+C=$
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