Question

How to integrate dx/(sin(x)+sin(2x)) ?

Answer 1

`I = 1/2ln|cosx + 1| + 1/6ln|cosx - 1| - 2/3ln|2cosx + 1| + C`

Explanation:

Using some trig:

`I = int 1/(sinx + 2sinxcosx)dx`

`I = int 1/(sinx(1 + 2cosx))dx`

`I = int sinx/(sin^2x(1 + 2cosx))dx`

`I = int (-sinx)/((cos^2x - 1)(1 + 2cosx)) dx`

`I = int (-sinx)/((cosx + 1)(cosx - 1)(2cosx + 1)) dx`

Now we let `u = cosx`. Then `du = -sinxdx` and `dx= (du)/(-sinx)`.

`I = int (-sinx)/((cosx + 1)(cosx - 1)(2cosx + 1)) * (du)/(-sinx)`

`I = int 1/((u + 1)(u - 1)(2u + 1)) du`

This becomes a partial fraction problem.

`A/(u + 1) + B/(u - 1) + C/(2u + 1) = 1/((u + 1)(u - 1)(2u + 1))`

`color(blue)(A(u - 1)(2u + 1) + B(u + 1)(2u + 1) + C(u + 1)(u - 1) =1`

`A(2u^2 - 2u + u - 1) + B(2u^2 + 2u + u + 1) + C(u^2 - 1) = 1`

`A(2u^2 - u - 1) + B(2u^2 + 3u + 1) + C(u^2 - 1) = 1`

`2Au^2 - Au - A + 2Bu^2 + 3Bu + B + Cu^2 - C = 1`

Thus

`{(2A + 2B + C = 0), (3B - A = 0), (B - A - C = 1):}`

If we substitute the second equation `A = 3B` into the first and third we get.
3...

`B - (3B) - C = 1`

`-2B - C = 1`

1...

`2(3B) + 2B + C = 0`
`6B + 2B + C = 0`
`8B + C = 0`

We can easily solve this system by elimination .

`6B = 1`
`B = 1/6`

Therefore

`8(1/6) + C = 0`

`C = -4/3`

And

`A = 3B = 3(1/6) = 1/2`

Thus the partial fraction decomposition is as follows:

`1/(2(u + 1)) + 1/(6(u - 1)) - 4/(3(2u + 1))`

`I=1/2int1/(u+1)du+1/6int1/(u-1)du-2/3intcolor(red)(2)/(color(red)((2u))+1)du`

This can be easily integrated.

`I = 1/2ln|u + 1| + 1/6ln|u - 1| - 2/3ln|2u + 1| + C`

subst. back , `u=cosx`

`I = 1/2ln|cosx + 1| + 1/6ln|cosx - 1| - 2/3ln|2cosx + 1| + C`

Hopefully this helps!

Note:
`A, B ,C` can obtain by putting `u=-1 , u=1 and u=1/2` into

BLUE EQUATION :

`color(blue)(A(u - 1)(2u + 1) + B(u + 1)(2u + 1) + C(u + 1)(u - 1) =1`

`u=-1=>A(-1-1)(-2+1)+B(0)+C(0)=1`

`=>A(-2)(-1)=1=>A=1/2`

`u=1=>A(0)+B(1+1)(2+1)+C(0)=1`

`=>B(2)(3)=1=>B=1/6`

`u=-1/2=>A(0)+B(0)+C(-1/2+1)(-1/2-1)=1`

`=>C(1/2)(-3/2)=1=>C=-4/3`