# Help trig integral question

1. Feb 26, 2008

### physstudent1

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

Evaluate the integral

integral of (tanx)^5*(secx)^4

2. Relevant equations

3. The attempt at a solution

what I did was broke the sec^4x into 2 sec^2's multiplied by each other
then one of these terms became tan^2x +1
so I had tan^5x*(tan^2x + 1)*sec^2x

then u = tanx
du = sec^2x

eventual answer tan^8x/8 + tan^6x/6 + c

however the book shows this problem by taking a (secxtanx) out of eveyrthing then doing tan^4x = (sec^2x -1)^2*sec^3x
u= secx
du = tanxsecx

eventual answer sec^8x/8 - sec^6x/3 + sec^4x/4 + C

...is there something wrong with the way I did it; I don't see anything wrong with the way I'm doing it but the book insists on doing all problems like this the otherway. I wouldn't mind but I don't like memorizing things and the way I always seem to think of doing these problems is the first method I wrote out please respond I have an exam on this stuff tomarrow night

2. Feb 26, 2008

### rocomath

$$\int\sec^{4}x\tan^{5}xdx$$

$$\int\sec^{3}x\tan^{4}x(\sec x\tan x)dx$$

$$\int\sec^{3}x(\sec^{2}x-1)^2(\sec x\tan x)dx$$

3. Feb 26, 2008

### physstudent1

yes that is how the book does it but why is the way that I did it wrong I don't get it?

4. Feb 26, 2008

### rocomath

Your way ...

$$\int\sec^{4}x\tan^{5}xdx$$

$$\int(\tan^{2}x+1)\tan^{5}x\sec^{2}xdx$$

I guess just have confidence in yourself. You did it all the steps correctly, but through a different route.

Last edited: Feb 26, 2008
5. Feb 26, 2008

### physstudent1

ok thanks alot; and wow thats a good way to proove it I didn't think of that

6. Feb 26, 2008

### rocomath

I take that back. I tried it and it didn't work, lol. But your solution is still right, but the limits of Integration is a different story.

7. Feb 27, 2008

### physstudent1

ah yea that does make sense that you would need different limits

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