Question

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

Answer 1

`f(x)=tanx-1/2cosx^2-x-sinx-2+1/2cospi^2+pi`

Explanation:

`f(x)=int(xsin(x^2)+tan^2x-cosx)dx=1/2int2xsin(x^2)dx+intsec^2xdx-int1dx-intcosxdx=-1/2cosx^2+tanx-x-sinx+"c"`

We are also told

`f(pi)=-1/2cospi^2+tanpi-pi-sinpi+"c"=-1/2cospi^2-pi+"c"=-2 rArr c=-2+1/2cospi^2+pi`

So

`f(x)=tanx-1/2cosx^2-x-sinx-2+1/2cospi^2+pi`

Answer 2

`f(x)=-1/2cosx^2+tanx-x-sinx+(pi-2-1/2cospi^2)`

Explanation:

Let
`I=int(xsinx^2+tan^2x-cosx)dx`
By sum rule

`I=int(xsinx^2)dx+int(tan^2x)dx+int(cosx)dx`
Let
`I_1=int(xsinx^2)dx`

`I_2=int(tan^2x)dx`

`I_3=int(cosx)dx`

`I=I_1+I_2-I_3`

Now,

`I_1=int(xsinx^2)dx`

Rearranging

`I_1=int(sinx^2)(xdx)`

Let
`u=x^2`

`sinx^2=sinu`

`(du)/dx=2x`

`xdx=1/2du`

`I_1=intsinu(1/2du)`

`I_1=-1/2int(-sinudu)`

`int-sinudu=cosu`

`u=x^2`

`I_1=-1/2cosx^2`

`I_2=int(tan^2x)dx`

`sec^2x-tan^2x=1`

`tan^2x=sec^2x-1`

`int(tan^2x)dx=int(sec^2x-1)dx`

`=intsec^2xdx-int1dx`

`intsec^2xdx=tanx`

`int1dx=x`

`int(tan^2x)dx=tanx-x`

`I_2=tanx-x`

`I_3=int(cosx)dx`

`intcosxdx=sinx`

`I_3=sinx`

`I=I_1+I_2-I_3`

`I=int(xsinx^2+tan^2x-cosx)dx`

`I_1=-1/2cosx^2`

`I_2=tanx-x`

`I_3=sinx`

`int(xsinx^2+tan^2x-cosx)dx=-1/2cosx^2+tanx-x-sinx`

Also

`f(pi)=-2`

Here,

`f(x)=int(xsinx^2+tan^2x-cosx)dx+C`

`f(pi)=-1/2cospi^2+tanpi-pi-sinpi+C`

`tanpi=0`

#sinpi=0

`-2=-1/2cospi^2-pi+C`#

`C=pi-2-1/2cospi^2`

`f(x)=-1/2cosx^2+tanx-x-sinx+(pi-2-1/2cospi^2)`