# Trig Integral help

• Feb 26th 2009, 09:17 PM
PHJ426
Trig Integral help
Indefinite integral of dx/(sin(x)+tan(x)

I replaced tan(x) with sin(x)/cos(x) so then I flipped the sinx/cosx term and wrote the integral as:

cosx/(sin(x) + sin(x) dx

becomes

cosx/(2 sinx) dx

Then I took the 1/2 out of the integral

1/2 times integral cos x / sin x dx

Let u = sin x du = cos x dx

1/2 times integral du / u

integrate

1/2 * ln u + C

replace sin x

1/2 * ln sin x + C

Is that correct?

And how do you do this with replacing sin x with 2u/ 1 + u^2 in the original problem and so on?

kind of hung up that way with the tangent function?

Thanks
• Feb 26th 2009, 09:25 PM
ThePerfectHacker
Quote:

Originally Posted by PHJ426
1/2 * ln sin x + C

Is that correct?

You should have absolute values.
And remember to check all you need to do is take the derivative and see if it works out!
• Feb 27th 2009, 04:43 AM
HallsofIvy
Quote:

Originally Posted by PHJ426
Indefinite integral of dx/(sin(x)+tan(x)

I replaced tan(x) with sin(x)/cos(x) so then I flipped the sinx/cosx term and wrote the integral as:

cosx/(sin(x) + sin(x) dx

There is a ")" missing in your first integral. Is this $\int \frac{dx}{\sin(x)+ \tan(x)}$?

If so then $\frac{1}{\sin(x)+ \tan(x)} = \frac{1}{\sin(x)+ \frac{\sin(x)}{\cos(x)}}$ and multiplying both numerator and denominator by cos(x) gives $\frac{\cos(x)}{\sin(x) \cos(x)+ \sin(x)}$, NOT what you have.

Quote:

becomes

cosx/(2 sinx) dx

Then I took the 1/2 out of the integral

1/2 times integral cos x / sin x dx

Let u = sin x du = cos x dx

1/2 times integral du / u

integrate

1/2 * ln u + C

replace sin x

1/2 * ln sin x + C

Is that correct?

And how do you do this with replacing sin x with 2u/ 1 + u^2 in the original problem and so on?

kind of hung up that way with the tangent function?

Thanks