If _- /2^ /28√2 xdx(1+e^ x)(1+ ^4x)= + _e(3+2√2), where , are integers, then ^2+ ^2 equals…

Question

If $\int _{-\pi /2}^{\pi /2}\frac{8\sqrt{2}\cos xdx}{(1+{e}^{\sin x})(1+{\sin }^{4}x)}=\alpha \pi +\beta {\log }_{e}(3+2\sqrt{2})$, where $\alpha ,\beta $ are integers, then ${\alpha }^{2}+{\beta }^{2}$ equals :

Answer: 8

Accepted Answer

$I=\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{8\sqrt{2}\cos x}{(1+{e}^{\sin x})(1+{\sin }^{4}x)}dx$ ............(1)

Apply king's rule

$I=\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{8\sqrt{2}\cos x({e}^{\sin x})}{(1+{e}^{\sin x})(1+{\sin }^{4}x)}dx$ ..........(2)

Adding (1) and (2), we get

$\begin{gathered} 2I=\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{8\sqrt{2}\cos x}{1+{\sin }^{4}x}dx \ I=\int _{0}^{\frac{\pi }{2}}\frac{8\sqrt{2}\cos x}{1+{\sin }^{4}x}dx, \end{gathered}$

Let $\sin x=t$

$\begin{gathered} I=8\sqrt{2}\int _{0}^{1}\frac{dt}{1+{t}^{4}} \ =4\sqrt{2}\int _{0}^{1}\frac{(1+\frac{1}{{t}^{2}})-(1-\frac{1}{{t}^{2}})}{{t}^{2}+\frac{1}{{t}^{2}}}dt \end{gathered}$

$\begin{gathered} =4\sqrt{2}\int _{0}^{1}\frac{(1+\frac{1}{{t}^{2}})dt}{{(t-\frac{1}{{t}^{2}})}^{2}+2}-4\sqrt{2}\int _{0}^{1}\frac{(1-\frac{1}{{t}^{2}})dt}{{(t+\frac{1}{{t}^{2}})}^{2}-2} \ =4\sqrt{2}\cdot \frac{1}{\sqrt{2}}(({{	an }^{-1}\frac{t-\frac{1}{t}}{\sqrt{2}}|}_{1}^{0}-4\sqrt{2}\cdot \frac{1}{2\sqrt{2}}{[\log |\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}|]}_{1}^{0} \ =2\pi -2\log |\frac{2-\sqrt{2}}{2+\sqrt{2}}| \end{gathered}$

$\begin{gathered} =2\pi +2\log (3+2\sqrt{2})=\alpha \pi +\beta {\log }_{e}(3+2\sqrt{2}) \ \Rightarrow \alpha =2,\beta =2 \ \Rightarrow {\alpha }^{2}+{\beta }^{2}=8 \end{gathered}$

Solutions · 1