How to integrate dx/(sin(x)+sin(2x)) ?
1 Answer
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
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))
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
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 ,
I = 1/2ln|cosx + 1| + 1/6ln|cosx - 1| - 2/3ln|2cosx + 1| + C
Hopefully this helps!
Note:
BLUE EQUATION :