Question

Find the integral of (1/2)(4/(1+sin(theta)))^2 d*theta from 0 to pi.

Answer 1

looks like youmight need tan(theta/2)=t

Answer 2

Can you please show the work?

Answer 3

build a right triangle for the sub I gave you

Answer 4

So we use u-substitution for u=tan^-1 t?

Answer 5

well i was using tan(theta/2)=t

Answer 6

I don't know how you got that in the first place.

Answer 7

also do you know your integral is improper?

Answer 8

Is it? How so?

Answer 9

wait whole it is to pi sorry
i thought it said 2pi

Answer 10

my substitution is commonly used to write trig functions as rationals
http://en.wikipedia.org/wiki/Tangent_half-angle_substitution

Answer 11

but the limits are weird
tan(0/2)=0=t
tan(t/2) goes to inf as t goes to pi
so this actually transforms it to an improper integral

Answer 12

an improper integral that will converge

Answer 13

let me know if you need any further help

Answer 14

tan(0/2)=0=t tan(t/2) goes to inf as theta goes to pi *

Answer 15

\[\frac{1}{2}\int_0^\pi\frac{4}{(1+\sin\theta)^2}~d\theta\]
There are at least two approaches to this sort of integral: the Weierstrass substitution that @myininaya provided the link for, and a method involving complex contour integration. Is this calculus or complex analysis?

Answer 16

This is Calculus 1.

Answer 17

And the 4 should be squared, it's (4/(1+sin theta))^2

Answer 18

Oh, in that case it might just be a matter of using some trig identities. It might not be the best way though.
\[\frac{1}{2}\int_0^\pi\left(\frac{4}{1+\sin\theta}\right)^2~d\theta\\
\frac{1}{2}\int_0^\pi16\left(\frac{1-\sin\theta}{(1+\sin\theta)(1-\sin\theta)}\right)^2~d\theta~~~~~\text{Note: this introduces a discontinuity}\\
8\int_0^\pi\left(\frac{1-\sin\theta}{1-\sin^2\theta}\right)^2~d\theta\\
8\int_0^\pi\left(\frac{1-\sin\theta}{\cos^2\theta}\right)^2~d\theta\\
8\int_0^\pi\frac{1-2\sin\theta+\sin^2\theta}{\cos^4\theta}~d\theta\\
8\int_0^\pi\left(\sec^2\theta-\frac{2\sin\theta}{\cos^4\theta}+\frac{\sin^2\theta}{\cos^4\theta}\right)~d\theta\]
The antiderivative for the first term is simply \(\tan\theta\), though you can leave it alone for now (you'll see why in a moment). The second requires a substitution, \(u=\cos\theta\). The third involves some intermediate steps before you can use a substitution:
\[\frac{\sin^2\theta}{\cos^4\theta}=\frac{1-\cos^2\theta}{\cos^4\theta}=\sec^4\theta-\sec^2\theta\]
This gives you
\[8\int_0^\pi\left(\sec^2\theta-\frac{2\sin\theta}{\cos^4\theta}+\sec^4\theta-\sec^2\theta\right)~d\theta\\
8\int_0^\pi\left(\sec^4\theta-\frac{2\sin\theta}{\cos^4\theta}\right)~d\theta\]
The first term, which I said you can ignore, disappears.
\[8\int_0^\pi\sec^4\theta~d\theta-16\int_0^\pi\frac{\sin\theta}{\cos^4\theta}~d\theta\]

Answer 19

Thanks a lot.