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

1 Answer
Aug 14, 2017

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!