Transcendentals

[tandoorichicken]

How do I perform this integral?
\int \sec^2{3x} \tan^5{3x} \,dx

[NateTG]

Originally posted by tandoorichicken
How do I perform this integral?
\int \sec^2{3x} \tan^5{3x} \,dx

Try u=\tan{3x}

[ShawnD]

Maple says the answer is
\frac{tan^6(3x)}{18}

Now lets see if we can get that.
First of all, the derivative of tan is sec^2. So that leads to me belive the process would be the substitution method.

I'll assign "U" as tan(3x)

\int sec^2(3x)tan^5(3x) dx

\int sec^2(3x)U^5 dx

Now take the derivative of U with respect to x

U = tan(3x)

\frac{dU}{dx} = 3sec^2(3x)

dx = \frac{dU}{3sec^2(3x)}

Now fill that back into what we had above

\int sec^2(3x)U^5 \frac{dU}{3sec^2(3x)}

\frac{1}{3} \frac{U^6}{6}

\frac{U^6}{18}

\frac{tan^6(3x)}{18}


Right on [:D]

[tandoorichicken]

Awesome thats what I got. Thanks guys.