Question

What is `f(x) = int xsinx^2 + secx dx` if `f(pi/12)=-8`?

Answer 1

The function is `f(x) = -1/2cos(2x) + ln|secx + tanx| - 8.62` approximately.

Explanation:

Start by separating the integrals.

`f(x) = int xsin(x^2) dx + int secx dx`

To solve the first integral, we let `u = x^2`. Then it follows that `du = 2xdx` and `dx = (du)/(2x)`.

`intxsin(x^2) = intxsinu(du)/(2x) = int1/2sinu du`

This integral is easily obtained.

`int 1/2sinu du = -1/2cosu + C = -1/2cos(x^2) + C`

Now to the second integral. This has already been derived by various sources as it is commonly encountered. The exact proof can be found here.

The integral is `intsecx dx = ln|secx + tanx| + C`

Putting all of this together, we obtain that

`f(x) = -1/2cos(x^2) + ln|secx + tanx| + C`

All that is left to do is to find the value of `C`. Using a calculator, you will find that `C ~~-8.62`.

Hopefully this helps!