1. ## 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

2. 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!

3. 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.

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