What is #f(x) = int xsinx^2 + sec^2x dx# if #f(0)=-1 #?
1 Answer
Sep 25, 2016
Explanation:
#f(x)=intxsin(x^2)dx+intsec^2(x)dx#
The second integral is simple: since
#f(x)=intxsin(x^2)dx+tan(x)+C#
For the remaining integral, use the substitution
#f(x)=1/2int2xsin(x^2)dx+tan(x)+C#
#f(x)=1/2intsin(u)du+tan(x)+C#
Since
#f(x)=-1/2cos(u)+tan(x)+C#
#f(x)=-1/2cos(x^2)+tan(x)+C#
Using the initial condition
#-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#