how to do this?

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[spare rose]

can i have other method other then get s2+c2=1
and t^(1/3)/(1+2sc)

i try u=pi/2-x on it but it doesn't work 😭

[zealous zinc]

Factor out the cosine in the denominator

Set $tan(x)=u^3$

[novel hawk]

😑 ɿototoЯ | Rototor 😑

[zealous zinc]

(Remember $\frac{1}{cos^2(x)}=1+tan^2(x)$)

[novel hawk]

😑 ɿototoЯ | Rototor 😑

[zealous zinc]

Then with some integration by parts you can simplify the integral to something easier

[fervent flint]

i wonder if theres some sort of reverse quotient rule use here

i give up integral calculator

integral calculator also gives up

[zealous zinc]

You can also directly integrate by parts but I think it’s easier to do so once the substitution is made

It basically all comes down to calculating $\int_{0}^{+\infty} \frac{1}{1+u^{3}} du$

[novel hawk]

😑 ɿototoЯ | Rototor 😑

[zealous zinc]

you need to know which steps to take to get there though I’ll let you see how to using what I told you

@spare rose

[fervent flint]

an idea i tried was to write sinx + cosx as a single function in terms of sin

and phase shift

perhaps this could be continued

[fervent flint]

new idea (NOT MINE) u = x - pi/4 which makes the integral symmetric about 0

okay but still the cbrt is so annoying 😭

[rotund minnow]

we could also use the gamma function

[fervent flint]

oh?

[rotund minnow]

yeah you just have to factor out a cos^2 from the denominator and youll see how it goes from there

[zealous zinc]

You can use the gamma function to evaluate $\int_{0}^{+\infty} \frac{1}{1+u^a} du$ for $a>1$ with a simple change of variables you can recognise the beta function, using the relation both functions have you can use Euler’s reflection formula

[novel hawk]

😑 ɿototoЯ | Rototor 😑

[zealous zinc]

Pretty neat tbh

[rotund minnow]

yeah its a nice integral

[limpid thicket]

partial fractions maybe

[spare rose]

thank you guys

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