Question

Integral cos^2 x sin^4 x dx

Answer 1

\[\int\limits_{}^{}\cos^2xsin^4x dx\]

Answer 2

Hmm they're both even powers, that's unfortunate.
We'll have to do a lot of work to solve this one.

We'll start by writing everything in terms of sine.
So make this change first \(\Large \cos^2x \quad=\quad 1-\sin^2x\) and multiply it out.

From there we can either apply `Half-Angle Formulas` a bunch of times.
Or we can jump right into the `Sine Reduction Formula`.
Have you learned about that yet?

Answer 3

\[\large \int\limits \sin^n x\;dx \quad=\quad -\frac{1}{n}\sin^{n-1}x \cos x +\frac{n-1}{n}\int\limits \sin^{n-2}x\;dx\]
Does this look familiar or no? :)

Answer 4

Definately not

Answer 5

Hmmm ok :(
Well it would certainly be easier but if we have to do Half Angles, then so be it. :P

Answer 6

\[\Large \int\limits \sin^4x-\sin^6x\;dx\]So ummm.

Answer 7

The Sine Half-Angle Identity:\[\Large \sin^2x \quad=\quad\frac{1}{2}(1-\cos2x)\]

Answer 8

So we can write sin^4 as,\[\Large \sin^4x\quad=\quad \left[\sin^2x\right]^2\quad=\quad \left[\frac{1}{2}(1-\cos2x)\right]^2\]

Answer 9

Then multiply it all out,\[\Large \frac{1}{4}(1-2\cos2x+\cos^22x)\]

Answer 10

Then apply the Cosine Half-Angle Identity to the last term there.\[\Large \frac{1}{4}(1-2\cos2x+\color{royalblue}{\cos^22x})\]
\[\Large =\frac{1}{4}(1-2\cos2x+\color{royalblue}{\frac{1}{2}(1+\cos4x)})\]

Answer 11

Now everything is written in terms of first powers. So they can be easily integrated, yay!

Answer 12

Hahaha! Do you multiply out first?

Answer 13

yes, distribute the fractions.\[\Large \int\limits \sin^4x-\sin^6x\;dx\]
That takes care of the sin^4x portion.
What to do about the sin^6x, hmmmm.

Answer 14

wait

Answer 15

The Equation is:\[\int\limits_{}^{}\cos^2xsin^4xdx\]

Answer 16

Oh you -_- you and your ways!

Answer 17

\[\large \int\limits \cos^2x \sin^4x\;dx\quad=\quad \int\limits (1-\sin^2x)\sin^4x\;dx \quad=\quad \int\limits \sin^4x-\sin^6x\;dx\]

Answer 18

Ah! Ok =) Now what can we do with sin^6?

Answer 19

hmmmm

Answer 20

hmmmmmmmmm

Answer 21

Is it possible to do like
\[\int\limits_{}^{} (\sin^2xcosx)^2\]
and apply a u-substitution?