compute integral

[janeann]

alright i have this to compute d/dx integral from 0 to tan(x). (theta^2+1)^3/2 d(theta) so far i have F'(x)=(x^2+1)^3/2=F'(tan(x))sec^2(x))-0 then by pythagorean i have sec^2(x) but i dont know what to do from their or if im even headed in the right direction.

 

[galactus]

It looks like you have the idea.

Second Fundamental Theorem of Calc.

\(\displaystyle \frac{d}{dx}\int_{a}^{g(x)}f(t)dt=f(g(x))g'(x)\)

\(\displaystyle \frac{d}{dx}\int_{0}^{tan(x)}(t^{2}+1)^{\frac{3}{2}}dt=\underbrace{(tan^{2}(x)+1)^{\frac{3}{2}}}_{\text{f(g(x))}}\cdot \overbrace{sec^{2}(x)}^{\text{g'(x)}}\)

But \(\displaystyle tan^{2}x+1=sec^{2}(x)\)

Simplifying gives:

\(\displaystyle sec^{3}(x)\cdot sec^{2}(x)=sec^{5}(x)\)