# U-sub trig integrals

1. Feb 11, 2009

### Knissp

1. The problem statement, all variables and given/known data
$$\int tan^5(3x) sec^2(3x) dx$$

3. The attempt at a solution

$$u = tan(3x)$$
$$du = 3 sec^2(3x) dx$$
$$du/3 = sec^2(3x) dx$$
$$\int tan^5(3x) sec^2(3x) dx$$
$$= \int 1/3 u^5 du$$
$$= u^6/18 + C$$
$$= tan^6(3x)/18 + C$$

The calculator says the integral should be:
$$\frac {1 - 3 sin^2(3x) cos^2(3x)}{18cos^6(3x)} + C$$

The answer I got does not differ from the calculator by a constant. Any help would be appreciated.

2. Feb 11, 2009

### Tom Mattson

Staff Emeritus
That's what I get too.

Yes it does. Sometimes it's tricky to show the equivalence (up to a constant) of two antiderivatives of trigonometric integrands. It frequently can depend on using just the right combination of identities to show that they are in fact the same. My advice is to put down the calculator and check your integration the old fashioned way: take the derivative of your result and verify that you recover the integrand.

3. Feb 11, 2009

### Knissp

My answer can be easily verified by differentiating.

$$d/dx \frac {1 - 3 sin^2(3x) cos^2(3x)}{18cos^6(3x)} + C$$
$$d/dx (\frac{1}{18cos^6(3x)} - d/dx \frac{3 sin^2(3x) cos^2 (3x)}{18 cos^6(3x)}$$
$$tan(3x) sec^6(3x) - d/dx \frac{sin^2(3x)}{6cos^4(3x)}$$
$$\frac {sin(3x)}{cos^7(3x)} - d/dx \frac{tan^2(3x) sec^2(3x)}{6}$$
$$\frac {sin(3x)}{cos^7(3x)} - \frac {tan^2(3x) d/dx (sec^2(3x)}{6} - \frac {sec^2(3x) d/dx tan^2(3x)}{6}$$
$$\frac {sin(3x)}{cos^7(3x)} - \frac {tan^2(3x) 2 sec(3x) (sec(3x)tan(3x)) 3}{6} - \frac {sec^2(3x) 2 tan(3x) sec^2(3x) 3}{6}$$
$$\frac {sin(3x)}{cos^7(3x)} - tan^2(3x) sec(3x) (sec(3x)tan(3x)) - sec^2(3x) tan(3x) sec^2(3x)$$
$$\frac {sin(3x)}{cos^7(3x)} - tan^3(3x) sec^2(3x) - sec^4(3x) tan(3x)$$
$$\frac {sin(3x)}{cos^7(3x)} - \frac{sin^3(3x)}{cos^5(3x)} - \frac{sin(3x)}{cos^5(3x)}$$
$$\frac {sin(3x)}{cos^7(3x)} - \frac{sin^3(3x) cos^2(3x)}{cos^7(3x)} - \frac{sin(3x)cos^2(3x)}{cos^7(3x)}$$

The integrand was $$tan^5(3x) sec^2(3x) = \frac{sin^5(3x)}{cos^7(3x)}$$, so all I need to do now is show
$$sin^5(3x) = sin(3x) - sin^3(3x) cos^2(3x) - sin(3x)cos^2(3x)$$

$$sin(3x) - sin^3(3x) cos^2(3x) - sin(3x)cos^2(3x)$$
$$sin(3x) - sin(3x)cos^2(3x) - sin^3(3x) cos^2(3x)$$
$$sin(3x) (1 - cos^2(3x)) - sin^3(3x) cos^2(3x)$$
$$sin(3x) (sin^2(3x)) - sin^3(3x) cos^2(3x)$$
$$sin^3(3x) - sin^3(3x) cos^2(3x)$$
$$sin^3(3x) (1 - cos^2(3x))$$
$$sin^5(3x)$$

cool! Thanks!!