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

1 Answer
Feb 23, 2018

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