What is F(x) = int sin2xcos^2x-sinxcos^2x dx if F(pi) = 1 ?

1 Answer
Jan 9, 2017

F(x) = (cos^3x)/3- (cos^4x)/2+11/6

Explanation:

Let's calculate the indefinite integral:

int (sin2xcos^2x-sinxcos^2x)dx = int cos^2x(sin2x-sinx)dx

Using the identity:

sin2x = 2 sinxcosx

int cos^2x(sin2x-sinx)dx = int cos^2x(2sinxcosx-sinx)dx=int cos^2xsinx(2cosx-1)dx

As sinxdx = -d(cosx)

int cos^2xsinx(2cosx-1)dx = int (cos^2x-2cos^3x)d(cosx) = (cos^3x)/3- (cos^4x)/2+C

Now we determine the constant from:

F(pi) = 1

(cos^3pi)/3- (cos^4pi)/2+C = 1

(-1)^3/3-(-1)^4/2 +C = 1

-1/3-1/2 +C =1

C = 1+5/6=11/6

So, finally:

F(x) = (cos^3pi)/3- (cos^4pi)/2+11/6