Anyone knows which method to use integration by parts or You-substitution?

[icy vale]

I literally tried doing this question by using integration by parts but I gave up halfway. Is there any other methods I could use beside U-substitution or integration by parts?

[crystal dome]

Do you know ||trig substitutions||

[icy vale]

Wut is that

@crystal dome you can use integration by parts right?

[crystal dome]

I would substitut $sin(u)^2=x$ and the integral becomes
$$\int \frac{2sin^4(u)cos(u)}{cos(u)}du= 2\int sin^4(u)du $$
Now you either use sine reduction formula or integration by parts. You should get:
$$ 2\int sin^4(u)du = \frac{3}{4}u- \frac{1}{2}sin(2u)+c=$$
And now you sub back and you can simplify

[fathom sparrow]

homelama

[icy vale]

Wah

Thanks

[hexed flame]

@crystal dome ? From what the other sine came from

[crystal dome]

the derivativ of $\left(sin^2(u)\right)‘=2sin(u)cos(u)$ and you multiply them

[fathom sparrow]

homelama

[icy vale]

Ye

Wait but how you know to sub in sin^2(u)

😭🙏

My brain is gone 😭

[hexed flame]

To Sqrt(1-x) become Sqrt(1-sin^2 u)=cos^2 u

[icy vale]

Ye

[hexed flame]

@crystal dome what divides the cosine in the bottom of the fraction? Sqrt(1-sin^2 u)=cos^2 u

[icy vale]

@crystal dome um so for u = sin^{-} \sqrt(x) right

[icy vale]

The square root divides it I think

[hexed flame]

Sorry

[crystal dome]

Here a full write up:
The starting integral is:
$$\int \frac{x^{\frac{3}{2}}}{\sqrt{1-x}}dx$$
At first we see that the fractional exponent is cinda strange so we substitut $v^2=x,2vdv=dx$:
$$\int \frac{v^{\frac{3}{2}\cdot 2}\cdot 2v}{\sqrt{1-v^2}}dv=\int \frac{2v^4}{\sqrt{1-v^2}}dv$$
Now we want to eliminate the square root. We know $sin^2(x)+cos^2(x)=1 \iff \sqrt{1-sin^2(x)}=cos(x)$. This is a pretty standard substitution but you have to get used to it: $sin(u)=v,cos(u)du=dv$:
$$\int \frac{2sin^4(u)cos(u)}{cos(u)}du= 2\int sin^4(u)du $$
Using the identity $$cos(2x)=cos^2(x)-sin^2(x) $$ $$\iff 1-cos(2x)=1+sin^2(x)-cos^2(x) $$ $$\iff 1-cos(2x)=sin^2(x)+cos^2(x)+sin^2(x)-cos^2(x) $$ $$\iff \frac{1-cos(2x)}{2}=sin^2(x)$$ Plugging this in our equation gives us:
$$\frac{1}{2} \int \left(1-cos(2u)\right)^2du=\frac{1}{2} \int 1-2cos(2u)+cos^2(2u)du$$
By a simalar idea we can rewrite $cos^2(2u)$
$$cos(2x)=cos^2(x)-sin^2(x) \iff 1+cos(2x)=1-sin^2(x)+cos^2(x) $$ $$\iff 1+cos(2x)=sin^2(x)+cos^2(x)-sin^2(x)+cos^2(x) $$ $$\iff \frac{1+cos(2x)}{2}=cos^2(x)$$
$$\frac{1}{2} \int 1-2cos(2u)+cos^2(2u)du=\frac{1}{2} \int 1-2cos(2u)+\frac{1+cos(4u)}{2}du=$$ $$\frac{3}{4}u-\frac{1}{2}sin(2u)+\frac{1}{16}sin(4u)+c$$
(I miscalculated eralier) Now we can substitut back $u=sin^{-1}\left(\sqrt{x}\right)$ and we get:
$$\frac{3}{4}u-\frac{1}{2}sin(2u)+\frac{1}{16}sin(4u)+c=$$ $$\frac{3}{4}sin^{-1}\left(\sqrt{x}\right)-\frac{1}{2}sin(2sin^{-1}\left(\sqrt{x}\right))+\frac{1}{16}sin(4sin^{-1}\left(\sqrt{x}\right))+c$$
You can for sure simplify the answer with trig identities but i am hinestly too lazy to do that.

[icy vale]

Oooo thank you

[fathom sparrow]

homelama

[teal copper]

Wow, that was very thorough.

[teal copper]

Don't forget to mark it as .solved when you're done 😁

[icy vale]

.close