What is f(x) = int xsinx^2 + tan^2x -cosx dx if f(0)=-1 ?

1 Answer
Jun 17, 2016

f(x)=-1/2cosx^2+tanx-x-sinx+1/2.

Explanation:

given that, f(x)=int[xsinx^2+tan^2x-cosx]dx=intxsinx^2dx+inttan^2xdx-intcosdx=I_1+I_2-I_3, say.

Then, I_1=intxsinx^2dx, =1/2intsinx^2*2xdx, =1/2intsintdt, [where x^2=t rArr 2xdx=dt]
=1/2(-cost)=-1/2cosx^2.

I_2=inttan^2xdx=int(sec^2x-1)dx=tanx-x.

I_3=intcosxdx=sinx.

f(x)=I_1+I_2-I_3=-1/2cosx^2+tanx-x-sinx+C.

But, given that f(0)=-1.

Hence, -1/2cos0+tan0-0-sin0+C=0. #:. C=1/2.

Finally,

f(x)=-1/2cosx^2+tanx-x-sinx+1/2.