What is f(x) = int xsinx^2 + secx dxf(x)=∫xsinx2+secxdx if f(pi/12)=-8 f(π12)=−8?
1 Answer
The function is
Explanation:
Start by separating the integrals.
f(x) = int xsin(x^2) dx + int secx dxf(x)=∫xsin(x2)dx+∫secxdx
To solve the first integral, we let
intxsin(x^2) = intxsinu(du)/(2x) = int1/2sinu du∫xsin(x2)=∫xsinudu2x=∫12sinudu
This integral is easily obtained.
int 1/2sinu du = -1/2cosu + C = -1/2cos(x^2) + C∫12sinudu=−12cosu+C=−12cos(x2)+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
Putting all of this together, we obtain that
f(x) = -1/2cos(x^2) + ln|secx + tanx| + Cf(x)=−12cos(x2)+ln|secx+tanx|+C
All that is left to do is to find the value of
Hopefully this helps!
