Question

Quick Integral Question:

(cos(x))^5 / sin(x)

The cos^5(x)...the x is not raised to the 5th the cosine is.... How would you find the integral of that. What rules should I use?

Answer 1

substitution

u = cos x
du = -sin x dx

giving
\[\int\limits \frac{u^5}{u^2 - 1} du\]

Answer 2

But the sine is in the denominator cow 0_o
I'm confused how you got that.

Answer 3

from here use long division
\[\rightarrow \int\limits u^3 + u + \frac{u}{u^2 - 1}\]
finally split up last fraction using partial fractions
\[\frac{u}{u^2 -1} = \frac{u}{(u+1)(u-1)} = \frac{A}{u+1} + \frac{B}{u-1}\]

Answer 4

@zepdrix,

dx = du/-sin yielding "-sin^2" in denominator which equals "cos^2 - 1"

Answer 5

Oh ok :) Interesting.

Here is how I would approach it personally.
Just another idea.

Break the cos^5 into cos^4 and cos,
write the cos^4 as cos^2 being squared,
\[\Large\rm \int\limits \frac{(\cos^2 x)^2 \cos x ~dx}{\sin x}\]Then apply your Pythagorean Identity:\[\Large\rm \int\limits\limits \frac{(1-\sin^2x)^2 \cos x ~dx}{\sin x}\]
Then let \(\Large\rm u=\sin x, \qquad du=\cos x~dx\)

Answer 6

so the denominator and the cosx cancel out leaving me with u^2?

Answer 7

@JohnathonNY , both methods will work but i would use zepdrix here, simpler steps

haha i over complicated it :)

Answer 8

no it would leave you with
\[\int\limits \frac{(1-u^2)^2}{u} du\]
expand, then divide each term by "u"