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# tan^3(x) integral in certain form?

Find the indefinite integral by factoring out tan x sec x and rewriting the integrand as powers of sec x multiplied by tan x sec x:

∫ tan^3(x) dx

### 2 Answers

- 9 years agoFavorite Answer
Haha I just did this problem for my roomie :D

So given the fact that secx will come into the problem, you'd want to use the identity sec^2(x) = tan^2(x) + 1 as adffg said.

Since tanx is raised to the power of three, factor one out so you'd have: ∫ tan(x)tan^2(x) dx

Now you can rewrite the integral using the identity: ∫ tan(x)(sec^2(x) -1) dx

Distribute the tan(x): ∫ [tan(x)sec^2(x) - tan(x) ]dx

Factor out the sec^2(x) so that you'd have the tan(x)sec(x) term: ∫ [tan(x)sec(x)sec(x) - tan(x)] dx

The question asks you to factor out the tan(x)sec(x): ∫ tan(x)sec(x) [sec(x) - (1/sec(x))] dx

Here you want to use u-substitution because tan(x)sec(x) is the derivative of sec(x). Therefore, make u = sec(x) and rewrite: ∫ (u - 1/u) du

Integrate with respect to u and you have: (u^2)/2 - ln(u) + C

Replace u with sec(x): (sec^2(x))/2 - ln(sec(x)) + C

There you go!

P.S. Gulick? or Berg?

- adffgLv 49 years ago
By trig indentity, Sec^2x = tan^2x + 1 so, tan^2x = sec^2x - 1

tan^3x = tan^2xtanx = (sec^2x-1)tanx

lets call A = ∫tan^3(x) dx

A = ∫(tan x)(sec2 x - 1)dx = ∫tanx sec^2x dx- ∫tanxdx = sec^2x/2 - ln|secx| + c