Question

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

Answer 1

`f(x)=-1/2cos(x^2)+tan(x)-1/2`

Explanation:

`f(x)=intxsin(x^2)dx+intsec^2(x)dx`

The second integral is simple: since `d/dxtan(x)=sec^2(x)`, we see that `intsec^2(x)dx=tan(x)+C`.

`f(x)=intxsin(x^2)dx+tan(x)+C`

For the remaining integral, use the substitution `u=x^2`. This implies that `du=2xdx`.

`f(x)=1/2int2xsin(x^2)dx+tan(x)+C`

`f(x)=1/2intsin(u)du+tan(x)+C`

Since `intsin(x)dx=-cos(x)+C`:

`f(x)=-1/2cos(u)+tan(x)+C`

`f(x)=-1/2cos(x^2)+tan(x)+C`

Using the initial condition `f(0)=-1`, we see that:

`-1=-1/2cos(0)+tan(0)+C`

`-1=-1/2(1)+0+C`

`C=-1/2`

Thus:

`f(x)=-1/2cos(x^2)+tan(x)-1/2`