Whats the step

I wanted to express

$\large{\displaystyle \int_{0}^{1} \frac{dx}{\sqrt{1+x^{4}}}}$ in terms of beta function

I started with $x=\sqrt{\tan t}$ , then got the integral in form of $sin2t$ and again substituted $sin2t=z$ , after this i couldn't solve

#Calculus

Easy Math Editor

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Comments

I assume you got to this point: $\displaystyle \frac 1{\sqrt2} \int_0^{\pi /4} (\sin(2t))^{-1/2} \, dt$ ,

Now let $y = 2t$ , you get $\displaystyle \frac1{2\sqrt2} \int_0^{\pi /2} (\sin(y))^{-1/2} \, dy$ .

Compare with equation 14 and equation 1 from this page.

And apply Euler Reflection Formula, you should get the answer of $\frac1{8\sqrt\pi} \left [ \Gamma \left( \frac 14\right)^2 \right ]$ .

Pi Han Goh - 6 years ago

Thanx for the help $\ddot \smile$

Tanishq Varshney - 6 years ago

@Azhaghu Roopesh M @Brian Charlesworth sir @Rajdeep Dhingra

Tanishq Varshney - 6 years ago

Sry couldn't make it in time . Btw are you participating in Proofathon ?

A Former Brilliant Member - 6 years ago

Nope, i am just a beginner to all this , not good as all other members on brilliant

Tanishq Varshney - 6 years ago

@Tanishq Varshney – No prob , just give it a try .Btw I am not able to solve 4 questions as of now .

A Former Brilliant Member - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$