Question

Integrate (cosx)^4. Is not it (1/4)(cosx)^4?

Answer 1

I messed up

Answer 2

Should be cosx ^3

Answer 3

It's not \(1/4 \cos(x)^4\) because the derivative of that is not \(\cos(x)^4\)

Answer 4

The derivative of \(\frac{1}{4}\cos^4(x)=4\cdot\frac{1}{4}\cos^3(x)\cdot (-\sin(x))\) by the "chain rule".

Answer 5

TY

Answer 6

\[\frac{d}{dx}\cos^4(x)\] or
\[\frac{d}{dx}\cos(x^4)\]?

Answer 7

The first one

Answer 8

you have to use some annoying "power reduction formula". look in the back of the cacl text

Answer 9

actually i missed a part, so ignore that one

Answer 10

Hey guys I messed up. Remember to double check before you ask a question, maybe 3 times. Better than failing.

Answer 11

\[\int\cos^n(x)dx = \frac{1}{n}\cos^{n-1}\sin(x)+\frac{n-1}{n} \int \cos^{n-2}(x)dx\] is the right one

Answer 12

That works too. Maybe not the simplest way.

Answer 13

so i guess you get
\[\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{4}\int \cos^2(x)dx\] and second one you do by rewriting
\[\cos^2(x)=\frac{1}{2}\cos(2x)+\frac{1}{2}\] and integrate term by term

Answer 14

hi joseph
try the identities also
intg(cos x)^4 dx=intg[(1+cos 2x)/2]^2 dx
=(1/4)intg[1+2cos 2x +(cos 2x)^2]dx
=(1/4)intg[1+2cos 2x +(1/2)(1+c0s 4x)]dx
=(1/4)intg[(3/2)+2 cos 2x+(1/2)(cos 4x)]dx
=(3/8)x+(1/4)sin 2x + (1/32) sin 4x +C ans...

Answer 15

i find the formulas easier since i find integrating this stuff stultifying. can never remember all the tricks other than integrating by parts, which is not really a trick, just backwards product rule.

Answer 16

yeah satellite is also correct you can also use integration by parts