help with these two integrals

[dreamy solar]

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[honest ruin]

For the first, let u=sqrt(x)

for the second, let u=x^2

both are trig integrals

$\int\sqrt{2\sqrt{x}-x}dx$, let $u=\sqrt{x}$, then we have $\int 2u\sqrt{2u-u^2}du$

[somber solstice]

vengeance

[honest ruin]

ill brb

[steady kindle]

no worries

for the second one i started

but idk this looks weird

so i stopped

[tawny linden]

u used 2 substitutions?

[honest ruin]

It's fine bruv

you will get the right answer if you sub everything back in

i mean you literally solved it

[steady kindle]

yeah subbing back in is the hard part imo

LOL

[steady kindle]

imma have to try this for the first one

brb

[honest ruin]

yeah this is pretty ezpz

$\int\sqrt{2\sqrt{x}-x}dx\overset{u=\sqrt{x}}{=}\int 2u\sqrt{2u-u^2}du=2\int u\sqrt{1-(u-1)^2}du\overset{v=u-1}{=}2\int (v+1)\sqrt{1-v^2}dv=2\int v\sqrt{1-v^2}dv+2\int\sqrt{1-v^2}dv=-\frac43(1-v^2)^{\frac32}+\arcsin(v)+v\sqrt{1-v^2}+C=-\frac43(1-(u-1))^{\frac32}+\arcsin(u-1)+(u-1)\sqrt{1-(u-1)^2}+C=\boxed{-\frac43(1-(\sqrt{x}-1))^{\frac32}+\arcsin(\sqrt{x}-1)+(\sqrt{x}-1)\sqrt{1-(\sqrt{x}-1)^2}+C}$

[somber solstice]

vengeance

[honest ruin]

lightwork no reaction

[honest ruin]

$\int\frac{x}{(1+4x^2+x^4)^{\frac52}}dx\overset{u=x^2}{=}\frac12\int\frac{1}{(1+4u+u^2)^{\frac52}}du=\frac12\int\frac{1}{((u+2)^2-3)^{\frac52}}du\overset{v=u+2}{=}\frac12\int\frac{1}{(v^2-3)^{\frac52}}dv\overset{v=\sqrt{3}\sec(\theta)}{=}\frac12\int\frac{1}{(3\sec^2(\theta)-3)^{\frac52}}\sqrt{3}\sec(\theta)\tan(\theta)d\theta=\frac12\int\frac{1}{3^{\frac52}}\cot^5(\theta)\sqrt{3}\tan(\theta)\sec(\theta)d\theta=\frac{1}{18}\int\cot^4(\theta)\sec(\theta)d\theta=\frac{1}{18}\int\frac{\cos^3(\theta)}{\sin^4(\theta)}d\theta=\frac{1}{18}\int\frac{(1-\sin^2(\theta)\cos(\theta)}{\sin^4(\theta)}d\theta\overset{w=\sin(\theta)}{=}\frac{1}{18}\int\frac{1-w^2}{w^4}dw=-\frac{1}{54}w^{-3}+\frac{1}{18w}+C=-\frac{1}{54}\csc^3(\theta)+\frac{1}{18}\csc(\theta)+C=-\frac{1}{54}\csc^3\qty(\arcsec(\frac{v}{\sqrt{3}}))+\frac{1}{18}\csc(\arcsec(\frac{v}{\sqrt{3}}))+C=-\frac{1}{54}\qty(\frac{v}{\sqrt{v^2-3}})^3+\frac{v}{18\sqrt{v^2-3}}+C=\boxed{-\frac{1}{54}\qty(\frac{x^2+2}{\sqrt{(x^2+2)^2-3}})^3+\frac{x^2+2}{18\sqrt{(x^2+2)^2-3}}+C}$

noooooooooooooo

one mfing } sign

ONE

@steady kindle

[somber solstice]

vengeance

[steady kindle]

alright ty

+close

[charred seal]

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