What is f(x) = int xsinx^2 + tan^2x -cos^2x dx if f(0)=-3 ?

1 Answer
Jun 6, 2018

f(x)=-1/2cosx^2+tanx-3/2x-1/2sinxcosx-5/2

Explanation:

f(x)=intxsinx^2+tan^2x-cos^2xdx

The first term is of the form g'(x)g(x), so integrates to -1/2cosx^2.

The second term we transform by use of a standard trig identity:
tan^2x=sec^2x-1. We know that d/dx(tanx)=sec^2x, so the second term integrates to tanx-x.

The third term we transform by use of another standard trig identity:
cos^2x=1/2(1+cos2x)
So
int-cos^2xdx=-1/2int1+cos2xdx=-1/2(x+1/2sin2x)
Recall that sin2x=2sinxcosx, and so the third term integrates to
-1/2(x+sinxcosx).

Put the three integrated terms together:
f(x)=-1/2cosx^2+tanx-3/2x-1/2sinxcosx+C

Solve for C using the given condition, f(0)=-3:
-3=-1/2+0-3/2*0-1/2*0*1+C
-3=-1/2+C
C=-5/2

So
f(x)=-1/2cosx^2+tanx-3/2x-1/2sinxcosx-5/2