Have i started this integral correctly

[vande060]

Homework Statement

S tan^5x*sec^4x dx

Homework Equations

The Attempt at a Solution

S tanx*(1-sec^2)^2 * sec^4x dx

u = secx
du = secxtanx dx

S (1-u^2)*u^3

S (1 -2u^2 + u^4)*u^3

S ( u^3 - 2u^5 + u^7)

1/4*u^4 - 1/3*u^6 + 1/8*u^8 + c

then of course substituting secx back in for u

S

 

[HallsofIvy]

Homework Statement

S tan^5x*sec^4x dx

Homework Equations

The Attempt at a Solution

S tanx*(1-sec^2)^2 * sec^4x dx

An error here: $tan^2(x)= sec^2(x)- 1$, not $1- sec^2(x)$
Remember that sin^2(x)+ cos^2(x)= 1 and you are dividing by cos^2(x) to get tan^2(x)+ 1= sec^2(x).

u = secx
du = secxtanx dx

because you can write the integral as
S (sec^2(x)- 1)^2 )(sec^3(x))(tan(x)sec(x)dx)
and those become
(u^2- 1)^2(u^3)(du)

S (1-u^2)*u^3

Three errors- you want u^2- 1, not 1- u^2, you forgot the square on that, and you forgot the "du".

Correcting those "typos", it will be completely correct and very clever!

S (1 -2u^2 + u^4)*u^3

S ( u^3 - 2u^5 + u^7)

1/4*u^4 - 1/3*u^6 + 1/8*u^8 + c

then of course substituting secx back in for u

S

 

[vande060]

great thanks, the typos come from a mad rush just before class started, thanks for looking through the errors to help me