# Thread: Trigonometric Integral

1. ## Trigonometric Integral

Hi. I'm working through some review questions in prep for the mid-term. This one has me scratching my head a little bit:

$\displaystyle \int{\frac{1}{1-sin(t)}}\,dt$

I would appreciate any hints. I'm not sure how to apply here the normal approach of using identies and substitution. I played around with a few techniques like adding one, and IBP, but this didn't seem to help.

2. ## Re: Trigonometric Integral

Hey infraRed.

One that comes to mind hint-wise is to multiply the inside by (1+sin(t))/(1+sin(t)) which will make the denominator 1 - sin^2(t) = cos^2(t) and this will simplify to sec^2(t) - tan^2(t) which is integrable using standard results.

3. ## Re: Trigonometric Integral

Hello, infraRed!

$\displaystyle \int \frac{1}{1-\sin t}\,dt$

Multiply by $\displaystyle \tfrac{1+\sin t}{1+\sin t}\!:\;\; \frac{1}{1-\sin t}\cdot\frac{1+\sin t}{1+\sin t} \;=\;\frac{1+\sin t}{1-\sin^2\!t} \;=\; \frac{1+\sin t}{\cos^2\!t}$

. . . . . . . . . . . $\displaystyle =\;\frac{1}{\cos^2\!t} + \frac{\sin t}{\cos^2\!t} \;=\;\sec^2\!t + \sec t\tan t$

And you can integrate: .$\displaystyle \int(\sec^2\!t + \sec t\tan t)\,dt$

Right?

4. ## Re: Trigonometric Integral

Originally Posted by Soroban
Hello, infraRed!

Multiply by $\displaystyle \tfrac{1+\sin t}{1+\sin t}\!:\;\; \frac{1}{1-\sin t}\cdot\frac{1+\sin t}{1+\sin t} \;=\;\frac{1+\sin t}{1-\sin^2\!t} \;=\; \frac{1+\sin t}{\cos^2\!t}$

. . . . . . . . . . . $\displaystyle =\;\frac{1}{\cos^2\!t} + \frac{\sin t}{\cos^2\!t} \;=\;\sec^2\!t + \sec t\tan t$

And you can integrate: .$\displaystyle \int(\sec^2\!t + \sec t\tan t)\,dt$

Right?
It might be easier for the OP to write \displaystyle \displaystyle \begin{align*} \int{\sec^2{(t)}\,dt} - \int{ \frac{-\sin{(t)}}{\cos^2{(t)}}\,dt} \end{align*} as there is a more obvious substitution.

5. ## Re: Trigonometric Integral

Originally Posted by Soroban
Hello, infraRed!

Multiply by $\displaystyle \tfrac{1+\sin t}{1+\sin t}\!:\;\; \frac{1}{1-\sin t}\cdot\frac{1+\sin t}{1+\sin t} \;=\;\frac{1+\sin t}{1-\sin^2\!t} \;=\; \frac{1+\sin t}{\cos^2\!t}$

. . . . . . . . . . . $\displaystyle =\;\frac{1}{\cos^2\!t} + \frac{\sin t}{\cos^2\!t} \;=\;\sec^2\!t + \sec t\tan t$

And you can integrate: .$\displaystyle \int(\sec^2\!t + \sec t\tan t)\,dt$

Right?
Thanks all! Forgot about the multiply-by-1 technique.

6. ## Re: Trigonometric Integral

An alternative method for dealing with integrals involving trig functions like this is to use the half-angle substitution

$\displaystyle u = \tan(t/2).$

That gets you $\displaystyle \tan(t)=\frac{2u}{1-u^{2}}, \quad \sin(t)=\frac{2u}{1+u^{2}}, \quad \cos(t) = \frac{1-u^{2}}{1+u^{2}},$ and $\displaystyle dt = \frac{2du}{1+u^{2}}.$

For this particular example it's not as neat as the routine already demonstrated, but there are times when it's the thing to do.

7. ## Re: Trigonometric Integral

Hint. Final answer should come to 2sin(t/2)/(cos(t/2)-sin(t/2)) + C

8. ## Re: Trigonometric Integral

Hint. Final answer should come to 2sin(t/2)/(cos(t/2)-sin(t/2)) + C