# Why Is My Answer Wrong (trig integral)

1. Feb 23, 2009

### moe darklight

agh, trig integrals are driving me nuts.

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

the integral of

tanx sec^4x dx

2. Relevant equations

3. The attempt at a solution

tanx sec^4x dx
= sec^3 tanx secx dx

u = secx
du = tanx secx dx

so,
= u^3 du
= (u/4)^4
= (1/4)sec^4x

2. Feb 23, 2009

### D H

Staff Emeritus
The final result looks good to me. You did make a mistake in an intermediate step, however; you should have had u^4/4, rather than (u/4)^4.

3. Feb 23, 2009

Yes, your final answer should be sec^4x/4, but you forgot one thing: +C. This is important.

4. Feb 23, 2009

### moe darklight

O, ok. The book substitutes u = tanx to get (1/4)tan^4x + (1/2)tan^2x + C, and that got me confused.

Thanks

5. Feb 24, 2009

### kof9595995

Just the same, since sec^2x=1+tan^2x

6. Feb 24, 2009

### HallsofIvy

The way I would have done this would have been to convert immediately to sine and cosine (I get so confused with those other trig functions!): tan(x) sec^4(x)= sin(x)/cos^5(x)[/itex] and then the substitution u= cos(x), du = -sin(x) dx gives the integral
[tex]-\int du/u^5= -\int u^{-5}du= (1/4)u^{-4}+ C= (1/4)cos^{-4}(x)+ C= (1/4)sec^4(x)+ C[/itex]
again, "just the same".