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

1 Answer
Sep 25, 2016

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