Integral involving sec and tan

[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 let's see if we can get that.
First of all, the derivative of tan is sec^2. So that leads to me believe 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

 

[tandoorichicken]

Awesome that's what I got. Thanks guys.