Trigonometric Integral

• Oct 31st 2013, 02:29 PM
infraRed
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:

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

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.
• Oct 31st 2013, 03:47 PM
chiro
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.
• Oct 31st 2013, 05:30 PM
Soroban
Re: Trigonometric Integral
Hello, infraRed!

Quote:

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

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

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

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

Right?
• Oct 31st 2013, 06:38 PM
Prove It
Re: Trigonometric Integral
Quote:

Originally Posted by Soroban
Hello, infraRed!

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

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

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

Right?

It might be easier for the OP to write \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.
• Oct 31st 2013, 06:45 PM
infraRed
Re: Trigonometric Integral
Quote:

Originally Posted by Soroban
Hello, infraRed!

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

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

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

Right?

Thanks all! Forgot about the multiply-by-1 technique.
• Nov 1st 2013, 02:23 AM
BobP
Re: Trigonometric Integral
An alternative method for dealing with integrals involving trig functions like this is to use the half-angle substitution

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

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

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.
• Nov 1st 2013, 11:57 PM
SelfTaughtMaths
Re: Trigonometric Integral
Hint. Final answer should come to 2sin(t/2)/(cos(t/2)-sin(t/2)) + C
• Nov 2nd 2013, 12:02 AM
SelfTaughtMaths
Re: Trigonometric Integral
Hint. Final answer should come to 2sin(t/2)/(cos(t/2)-sin(t/2)) + C