# Help trig integral question

## Homework Statement

Evaluate the integral

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

## 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

Related Calculus and Beyond Homework Help News on Phys.org
$$\int\sec^{4}x\tan^{5}xdx$$

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

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

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

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

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

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

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

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

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