Question

\[\int\limits_{0}^{10} \frac{ 1 }{ 3+2 \sin x } dx\]
Halp.
Can someone give me a step-by-step explanation on how to do this without a graphing calc?

Answer 1

Hmm I've seen this one a few times.
I know it involves some kind of weird substitution like u=tan(x/2) but I'm not sure how to work it.

I'm gonna follow the steps on wolfram and see if I can make sense of it.
I'm pretty sure harttn would know how to do this one.

Answer 2

Alright, I guess i'll leave it up for now then.

Answer 3

That's what I tried, I'm playing around with it right now, pretty fun! =)

Answer 4

@zepdrix mentioned the right substitution.

If \(u=\tan\dfrac{x}{2}\), then \(du=\dfrac{1}{2}\sec^2\dfrac{x}{2}~dx\), or \(2\cos^2\dfrac{x}{2}~du=dx\).

Now, from this substitution we have the following relations:
\[\sin x=\sin\left(2\cdot\frac{x}{2}\right)=2\sin\frac{x}{2}\cos\frac{x}{2}\\
\cos x=\cos\left(2\cdot\frac{x}{2}\right)=\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\]
You can draw yourself a triangle to see how you can write these relations in terms of \(u\):

\[\sin\frac{x}{2}=\frac{u}{\sqrt{u^2+1}}\\
\cos\frac{x}{2}=\frac{1}{\sqrt{u^2+1}}\]
And so now,
\[\sin x=\frac{2u}{u^2+1}\\
\cos x=\frac{1-u^2}{u^2+1}\]
Also, you get \(dx=\dfrac{2}{u^2+1}~du\).

Now on to the integral:
\[\begin{align*}\int\frac{dx}{3+2\sin x}&={\Large\int}\frac{\dfrac{2}{u^2+1}}{3+2\cdot\dfrac{2u}{u^2+1}}du\\
&={\Large\int}\frac{\dfrac{2}{u^2+1}}{\dfrac{3(u^2+1)+4u}{u^2+1}}du\\
&=\int\frac{2}{3(u^2+1)+4u}du\\
&=\int\frac{2}{3u^2+4u+1}du
\end{align*}\]
The remaining integral can be done by completing the square in the denominator, then applying a trig substitution.

There might be an error somewhere in the mix, but the idea is there.