Question

integration question
how would I find
indefinate intergal (tanx)^5(secx)^3 dx

Answer 1

calculator

Answer 2

jk do some tricks with your trig knowledge

Answer 3

what trig knowledge?

Answer 4

lol I'll give it a try --I'm assuming you mean trig identities?

Answer 5

yeah, you know tanx =sinx/cosx
you also know secx=1/cos

Answer 6

you also know that d/dx(tanx)=sec^2(x)

Answer 7

I do now!

Answer 8

You should use \( \frac{d}{dx} (secx)=tanx~secx \)

Answer 9

derivative of secx=tanxsecx

Answer 10

yeah

Answer 11

sin^2(x)=1-cos^2(x)
I solved it this way

Answer 12

\[\int\limits \tan^5(x)\sec^3(x) dx\] - make the powers even
\[\int\limits \tan^4(x)\sec^2(x)\sec(x)\tan(x)dx\]
let u=sec(x), therefore du=sec(x)tan(x)dx
\[\int\limits (\tan^2(x))^2\sec^2(x)\sec(x)\tan(x)dx = \int\limits (\sec^2(x)-1)^2\sec^2(x)\sec(x)\tan(x)dx\]
\[= \int\limits (u^2-1)^2u^2du = \int\limits (u^4-2u^2+1)u^2du = \int\limits (u^6-2u^4+u^2)du\]
Integrate
\[(u^7/7)-(2u^5/5)+(u^3/3)+C\]
Replace u with sec(x), then you're done

Answer 13

Another method: Convert everything to cos(x) and set u=cos(x)
\(\rm \int tan^5(x)sec^3(x)dx\)
\(\rm= \int sin^4(x)/cos^8(x)~~sin(x)dx\)... set u=cos(x), du=-sin(x)dx
\(\rm= \int -(1-u^2)^2/u^8 du\)
\(\rm= \int -(1/u^8-2/u^6+/u^4) du\)
\(\rm= 1/(7u^7)-2/(5u^5)+1/(3u^3) + C\) .... here u=cos(x)
as @ShaeWX had it.

Answer 14

trig was never my strong suit. Thanks guys

Answer 15

I'm going to keep working on it until is sinks in

Answer 16

@pblaesre
That's the best way! :)