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

Question

1263 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-25T19:03:23

#f(x)=-1/2cos(x^2)+tan(x)-1/2#

#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#