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?
Best Answer
$\displaystyle J=\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$
Perform the change of variable $y=\dfrac{1-x^4}{1+x^4}$,
$\displaystyle J=\dfrac{1}{2^{\tfrac{9}{4}}}\int_0^1 \dfrac{x^{\tfrac{3}{4}}}{(1-x)^{\tfrac{3}{4}}}dx=\boxed{\dfrac{1}{2^{\tfrac{9}{4}}}\text{B}\left(\dfrac{1}{4},\dfrac{7}{4}\right)}$