# The Same Integral Two Different Answers

1. Jun 10, 2007

### alba_ei

The Same Integral... Two Different Answers!!!

1. The problem statement, all variables and given/known data

$$\int \frac{\tan x}{\cos^5 x} \,dx$$

3. The attempt at a solution

Solution 1
$$\int \frac{\tan x}{\cos^5 x} \,dx = \int \frac{\sin x}{\cos^6 x}$$

$$u = \cos x \,\,\,\, du = -\sin x \,dx$$

$$= -\int u^-^5 \,du = \frac{1}{4 u^4} + C$$

Answer 1: $$\frac{1}{4 \cos^4 x} + C$$

Solution 2

$$\int \frac{\tan x}{\cos^5 x} \,dx = \int \tan x \sec^4 x \,dx = \int \tan x (1 + \tan^2 x) \sec^2 x \,dx$$

$$u = \tan x \,\,\,\, du = \sec^2 x \,dx$$

$$= \int \(u + u^3) \,du = \frac{1}{2} u^2 + \frac{1}{4} u^4 + C$$

Answer 2: $$\frac{1}{2} \tan^2 x + \frac{1}{4} \tan^4^ x + C$$

Wich one is right and why?

Last edited: Jun 10, 2007
2. Jun 10, 2007

### durt

In the first step of your second solution, you should have $$\sec^5{x}$$, not $$\sec^4{x}$$. You're also off by a power in your first solution. Close, though.

P.S.: Check your answers by taking derivatives and see if you arrive back at the integrand.

3. Jun 10, 2007

### Kurdt

Staff Emeritus
In the first attempt you should have $$-\int u^-^6 \,du$$

And similarly in the second attempt you have replaced $$\frac{1}{cos^5(x)}$$ with sec to the power 4 which is not correct.

4. Jun 10, 2007

### neutrino

I will tell you that neither of the answers are right. You find where the erroneous step is.(It's quite simple, really)
[EDIT]Hmm...never mind.

A further simplification of the second method is to write down the integral as
$$\int \frac{\tan x}{\cos^5 x} \,dx = \int \tan x \sec^5 x \,dx = \int \tan x \sec x \sec^4 x \,dx$$

5. Jun 10, 2007

### Gib Z

Whenever you seem to get two solutions, differentiate both of them. If they both arrive back at the integrand, then try and see why the 2 solutions have a difference of a constant, thats what the +C's there for = ]