The Same Integral Two Different Answers

[alba_ei]

The Same Integral... Two Different Answers!

Homework Statement

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

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?

 

[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.

 

[Kurdt]

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.

 

[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

 

[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, that's what the +C's there for = ]