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Vi Nguyen · May 7, 2009, 07:52 AM

Can someone tell me how to evaluate the integral:

∫tan(2x)sec³(2x)dx

I don't know what to do when sec is cubed and which substitution to make, or if there even is a substitution, if I express sec³(2x) as sec²(2x)sec(2x) and change sec²(2x) to (tan²(2x)+1), then let u=tan(2x) du/dx=sec²(2x) but I only have a sec(2x) left to cancel, :(

galactus · May 7, 2009, 08:25 AM

$\int tan(2x)sec^{3}(2x)dx$

Make the sub:

$u=2x, \;\ \frac{1}{2}du=dx$

$\frac{1}{2}\int tan(u)sec^{3}(u)du$

Now, make the sub $w=sec(u), \;\ dw=sec(u)tan(u)du$

See? The dw takes up everything but $sec^{2}(u)$ , which we use w^2 on

So, we have:

$\frac{1}{2}\int w^{2}dw$

Integrate:

$\frac{1}{6}w^{3}$

Resub:

$\frac{1}{6}sec^{3}(u)$

resub:

$\frac{1}{6}sec^{3}(2x)+C$

The C is a hanger on. Add it if the professor is anal.