Effortlessly Evaluate Integral Involving Sec with Limits -pi/3 to 0

[kuahji]

evaluate

$$\int2(sec x)^3$$ with the limits as -pi/3 to 0

I tried all sorts of things from breaking it apart to substitution, but known of what I tried work.

The book shows setting u=sec x & v=tan x

Then it shows the first step as 2 (sec x tan x) - 2 $$\int(sec x) * (tan x)^2 dx$$ then evaluate both parts to -pi/3 to 0.

Which is really what I'm not understanding. How did they integrate the first part & then still have the next part? I'm also not seeing how u & v come into play.

Guess I'm just plain lost on this one.

 

[alfredska]

Substitution

As Griffith's puts it, paraphrased, you can move the derivative from one variable to the other under an integral, and you'll just pick up a minus sign and a boundary term.

Thus the equation:
$$\int_a^buv'dx=\left.uv\right|_a^b-\int_a^bu'vdx$$

 

[rocomath]

$$\int\sec x(\tan^{2}x+1)dx$$
$$\int\sec x\tan^{2}xdx+\int\sec xdx$$

$$u=\sec x$$
$$du=\sec x \tan x dx$$

$$dV=\tan^{2}xdx$$
$$V=\sec x$$