Trig Integrals and Derivatives

[crayzeerunner]

I have a question abuot Intergrals of Trigfunctions.

My first question is, I know Intergral(sec^2u)du= tanu + C . So does this mean Intergral(tanu) = Sec^2u + C or not? I dont think that it does but that is the only formula I have containing tangent with an intergral and the problem I am trying to complete is

Intergral(tanx^4)(Secx^2)dx

The only insight I have on this problem is that Intergral of sec^2x = tanu + C. Can anyone help me? Thanks.

 

[stapel]

What is the derivative of sec²(x)?

What is the derivative of tan(x)?

Eliz.

 

[skeeter]

first off ... it is spelled integral, not intergral

No ... the antiderivative of tan(x) is not sec²x.

INT tan(x) dx = INT sin(x)/cos(x) dx = -INT -sin(x)/cos(x) dx = -ln|cos(x)| + C =
ln|sec(x)| + C


for INT tan⁴(x) sec²(x) dx, use the method of substitution ...
let u = tan(x), du = sec²(x) dx
you should get (1/5)tan⁵(x) + C