Question

∫ 7tan^5xdx

Answer 1

\[\int\limits7 \tan^5x dx\]

Answer 2

Turn it into
\[\int\limits_{}^{}7(\tan ^{2}x)(\tan ^{2}x)(tanx)dx\]Then take the two tan^2x terms and turn them into secants using the identity:
\[\tan ^{2}x + 1 = \sec ^{2}x \]After that you should be able to multiply things out and integrate.

Answer 3

Ok, I got that far but my final answer is wrong

Answer 4

my final answer was
7/4 sec^4x- tan^2x + ln Isec xI

Answer 5

Yeah, looks like we might have to try it only converting one of the tan^2's
\[7\int\limits_{}^{}(\sec ^{2}x-1)(\sec ^{2}x-1)(tanx)\]
\[7\int\limits_{}^{}(\sec ^{4}x-2\sec ^{2}x-1)(tanx)dx\]
\[7[\int\limits_{}^{}\sec ^{4}xtanxdx-2\int\limits_{}^{}\sec ^{2}xtanxdx-\int\limits_{}^{}tanxdx]\]
The first two integrals let u = secx
u = secx, du = secxtanxdx, dx = du/(secxtanx)
\[7[\int\limits_{}^{}u^{4}tanx*\frac{ du }{ utanx }-2\int\limits_{}^{}u ^{2}tanx*\frac{ du }{ utanx }-\int\limits_{}^{}tanxdx\]

Answer 6

the tanx's cancel and one u cancels in each one. So youd have u^3 for the firstone, u^1 for the second one. The derivative of tanx = -ln|secx|

Answer 7

is what you wrote the final answer?

Answer 8

*integral of tanx
my bad. And I wrote out everything but the actual integration of the u's and the tanx.

Answer 9

that looks confusing. Is there a way to simplify this more?

Answer 10

\[7[\int\limits_{}^{}u ^{3}-2\int\limits_{}^{}u-\int\limits_{}^{}tanx]\]
Once I cancel things out I have that. Without actually integrating, that's as simple as I can get it down to.

Answer 11

ok

Answer 12

I still don't see how my answer was wrong then....

Answer 13

Hmm...
\[7(\frac{ u ^{4} }{ 4 }-u ^{2}-\ln|secx|)\]

\[\frac{ 7\sec ^{4}x }{ 4 }-7\sec ^{2}x-7\ln|secx|+C\]

Answer 14

ok, i didnt have the 7's in front of all those

Answer 15

And you had a different sign in front of ln|secx|

Answer 16

it says that is right. thank you so much!

Answer 17

oh yea! i did

Answer 18

Awesome, glad ya got it ^_^