Question

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

Answer 1

`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.`