trig sub

[jade zephyr]

I'm kind of confused on how to proceed with, I'm in the middle of it, i know to have (1-cos^2(theta))^2*sin(theta)/ cos^2(theta) and to u-sub but im not sure where else to go from here

[glad frigate]

$$\int \frac{x^5}{\qty(4x^2 + 4)^{\frac 52}} \dd{x}$$

[versed stag]

King Leo

[warped zephyr]

x=tan(u) will do the trick 99% sure

From some napkin calculations, x=tan(u) will indeed work with just some simple trigonometry after

[glad frigate]

You seem to have done x = sin(theta), but that wont work here

[jade zephyr]

oh i know its tan(theta) im just confused in the middle of the process. especially with the added 4 in front. i tried to do it but it led me to 1/32 tan^5(theta)/sec^3(theta) and things got more confusing from there

[warped zephyr]

hint:

||sin^2=1-cos^2||

and then

break up the integral and use

||tan^2=sec^2-1||

[jade zephyr]

im cautious on the split integral with the u's. like the u^2/u^2 is that 1? or the u^4/u^2

[warped zephyr]

$\frac{1}{32}\int \sin^3(u)\tan^2(u)du=\frac{1}{32}\int \sin(u)(1-\cos^2(u))\tan^2(u)=\frac{1}{32}\left(\int \sin(u)\tan^2(u)du-\int \sin^3(u)du\right)$

[versed stag]

trigonometria

[warped zephyr]

in the first integral you can substitute cos(t) and the rest is smooth.

[jade zephyr]

im still kinda lost..

[warped zephyr]

Where exactly? Lets work on it together

[jade zephyr]

so from what you mentioned im at -1/32 tan^2t (1-u^2) do i multiply the two together? cos is now u so it'd just be -tan^2t*u^2

[warped zephyr]

Lets start over

Under $x=\tan(u)$ we have $$I=\frac{1}{32}\int \sin^3(u)\tan^2(u)du$$ Notice that, $\sin^3(u)=\sin^2(u)\sin(u)=\sin(u)(1-\cos^2(u))$ therefore $$I=\frac{1}{32}\int\sin(u)(1-\cos^2(u))\tan^2(u)du=\frac{1}{32}\int\sin(u)\tan^2(u)du-\frac{1}{32}\int\sin^3(u)$$ For the integrand $\sin(u)\tan^2(u)$ we can use the identity $\tan^2(u)=\sec^2(u)-1$ thus , $$I=\frac{1}{32}\int \sin(u)(\sec^2(u)-1)du-\frac{1}{32}\int\sin^3(u)du$$ Then, in the first integral we can substitute $s=\cos(u)$ and get $$I=-\frac{1}{32}\int\frac{1}{s^2}-1 \space ds -\frac{1}{32}\sin^3(u)du=\frac{1}{32s}+\frac{s}{32}-\frac{1}{32}\int\sin^3(u)du$$

[versed stag]

trigonometria

[warped zephyr]

Then use the reduction formula to get $$\int\sin^m(u)du=-\frac{\cos(u)\sin^{m-1}(u)}{m}+\frac{m-1}{m}\int\sin^{-2+m}(u)du$$ For $m=3$ and the rest is simple

[versed stag]

trigonometria

[jade zephyr]

oh i was way off..