Evaluate: $\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$

Question

Here is what I did:

$\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$

Put $x^2=\tan (\theta )$

Then: $x=\sqrt{\tan (\theta )}$ and $dx=\frac{1}{2} \tan ^{-\frac{1}{2}}(\theta ) \sec ^2(\theta )d\theta$

So the integral becomes $\int_0^{\frac{\pi }{4}} \frac{\left(1-\tan ^2(x)\right)^{3/4} \tan ^{-\frac{1}{2}}(x) \sec ^2(x)}{2 \left(\tan ^2(x)+1\right)^2} \, dx$

= $\frac{1}{2} \int_0^{\frac{\pi }{4}} \frac{\sin ^{-\frac{1}{2}}(x) {\cos^{\frac{1}{2}} (x)} \sec ^2(x) \left(\frac{\cos ^2(x)-\sin ^2(x)}{\cos ^2(x)}\right)^{3/4}}{\sec ^4(x)} \, dx$

= $\frac{1}{2} \int_0^{\frac{\pi }{4}} \sin ^{-\frac{1}{2}}(x) \cos (x) \cos ^{\frac{3}{4}}(2 x)\, dx$

Now how do I proceed?

I tried converting it to the form $\sin ^p(2 x) \cos ^q(2 x)$ but that proved impossible.

The textbook gives the answer as $\frac{1}{2^{9/2}}B\left(\frac{7}{4},\frac{1}{4}\right)$ which WolframAlpha tells me is 0.1472... but if ask it to evaluate the integral, I get the answer as 0.700... Who is right?

Answer 1

These are three successive substitutions that will get you to the answer:

First, use: $ a) \, x^2 = \tanh{\theta}$

Then, use: $ b) s^2 = \sinh{\theta}$

Finally, use: $ c) \, u = 2s^4$

Then use the following definition of the Beta Function:

$\displaystyle \text{B}(p,q) = \int_0^\infty \frac{x^{p-1}}{(1+x)^{p+q}} \text{d}t$

The answer is $\displaystyle \frac{1}{4\sqrt[4]{2}}\text{B}\left(\frac{1}{4},\frac{7}{4}\right) = \frac{3\pi \sqrt[4]{2}}{16}$

Answer 2

$\displaystyle J=\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$

Perform the change of variable $y=\dfrac{1-x^4}{1+x^4}$,

$\displaystyle J=\dfrac{1}{2^{\tfrac{9}{4}}}\int_0^1 \dfrac{x^{\tfrac{3}{4}}}{(1-x)^{\tfrac{3}{4}}}dx=\boxed{\dfrac{1}{2^{\tfrac{9}{4}}}\text{B}\left(\dfrac{1}{4},\dfrac{7}{4}\right)}$