Integration with Trigonometry

Relevant wiki: Integration U-substitution - Given U

$\mathcal I= \int_{\frac{\sqrt{2}}{\sqrt[8]{3}}}^{\sqrt{2}}\frac{4x^{3}}{(x^{4})^2-4} dx$ Substitute $x^4=t$ such that $4x^3 dx =dt$ :

$\mathcal I= \int_{\frac{4}{\sqrt 3}}^4\dfrac{dt}{t^2-4}\\=\dfrac 14\left( \int_{\frac{4}{\sqrt 3}}^4\dfrac{dt}{t-2}- \int_{\frac{4}{\sqrt 3}}^4\dfrac{dt}{t+2}\right)$

$\large =\left[\ln\left(\dfrac{t-2}{t+2}\right)^{1/4}\right]_{\frac{4}{\sqrt 3}}^4$

$\large =\ln\left(\dfrac{(2+\sqrt 3)^2}{3}\right)^{1/4}$ $\large =\ln\sqrt{\left(\dfrac{2+\sqrt 3}{\sqrt 3}\right)}$

Hence, $\large 2+3=\boxed 5$

Let $x=\sqrt[4]{2\sec \theta}$

$\frac{\mathrm{d}x }{\mathrm{d}\theta}=\frac{1}{4}(2\sec\theta)^{-\frac{3}{4}}(2\sec\theta \tan\theta)=\frac{2\sec\theta \sqrt{\sec^{2}\theta-1}}{4(2\sec\theta)^{\frac{3}{4}}}=\frac{\sec\theta\sqrt{4\sec^{2}\theta -4}}{4(2\sec\theta)^{\frac{3}{4}}}$

$\mathrm{d}x=\frac{\sec\theta\sqrt{4\sec^{2}\theta -4}}{4(2\sec\theta)^{\frac{3}{4}}}\mathrm{d}\theta$

Changing bounds

$\sqrt{2}=\sqrt[4]{2\sec \theta}$

$4=2\sec\theta$

$\cos\theta=\frac{1}{2}$ so $\theta=\frac{\pi}{3}$

${\frac{\sqrt{2}}{\sqrt[8]{3}}}=\sqrt[4]{2\sec \theta}$

$\frac{4}{\sqrt{3}}=2\sec\theta$

$\cos\theta=\frac{\sqrt{3}}{2}$ so $\theta=\frac{\pi}{6}$

Now we have

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{4(2\sec\theta)^{\frac{3}{4}}}{4\sec^{2}\theta -4}\times \frac{\sec\theta\sqrt{4\sec^{2}\theta -4}}{4(2\sec\theta)^{\frac{3}{4}}}\mathrm{d}\theta=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{2\sec\theta \tan\theta}{4\tan^{2}\theta}\mathrm{d}\theta$

$=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sec\theta}{\tan\theta}\mathrm{d}\theta=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\csc\theta \mathrm{d}\theta$

To solve we can multiply top and bottom by $(\csc\theta +\cot\theta)$

$=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\csc\theta (\csc\theta +\cot\theta)}{ \csc\theta +\cot\theta} \mathrm{d}\theta=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\csc^{2}\theta +\csc\theta \cot\theta}{\csc\theta +\cot\theta} \mathrm{d}\theta$

Using u-substitution

$u=\csc\theta + \cot\theta$

$\frac{\mathrm{d}u}{\mathrm{d}\theta}=-\csc\theta \cot\theta -\csc^{2}\theta$

$\mathrm{d}\theta=-\frac{1}{\csc\theta \cot\theta +\csc^{2}\theta}\mathrm{d}u$

$\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\csc^{2}\theta +\csc\theta \cot\theta}{u} \times \ (-\frac{1}{\csc\theta \cot\theta +\csc^{2}\theta})\mathrm{d}u=-\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{u}\mathrm{d}u$

$=\left [ -\frac{1}{2}\ln\left | \csc\theta +\cot\theta\right | \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$

$-\frac{1}{2}\ln \left | \frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}} \right |+\frac{1}{2}\ln \left | 2+\sqrt{3} \right|$

$\frac{1}{2}(\ln(2+\sqrt{3})-\ln(\sqrt{3}))=\frac{1}{2}\ln \left (\frac{2+\sqrt{3}}{\sqrt{3}} \right)=\ln \sqrt{\frac{2+\sqrt{3}}{\sqrt{3}}}$

$\therefore A+B=2+3=\mathbf{5}$