Evaluar

Question

Evaluar:% $$I = \int_{0}^{\Large\frac\pi2} (\sqrt{\sin x}+\sqrt{\cos x})^{-4}\ dx$$

Intento: \begin{align} I&=\int_{0}^{\Large\frac\pi2} (\sqrt{\sin x}+\sqrt{\cos x})^{-4}\ dx\\ &=\int_{0}^{\Large\frac\pi2} \left( \frac{\sqrt{\sin x}-\sqrt{\cos x}}{\sin x - \cos x}\right)^{4}\ dx \end {align}

¿Es la forma correcta de proceder es? Por favor guíame. Gracias.

Answer 1

\begin{align} \int_0^{\Large\frac\pi2} (\sqrt{\sin x}+\sqrt{\cos x})^{-4}\ dx&=\int_0^{\Large\frac\pi2}\frac1{(\sqrt{\cos x})^4\ (1+\sqrt{\tan x})^4}\ dx\\ &=\int_0^{\Large\frac\pi2}\frac{\sec^2x}{(1+\sqrt{\tan x})^4}\ dx\\ &\stackrel{\color{red}{[1]}}=\int_0^\infty\frac{du}{(1+\sqrt{u})^4}\\ &\stackrel{\color{red}{[2]}}=2\int_0^\infty\frac{t}{(1+t)^4}\ dt\\ &\stackrel{\color{red}{[3]}}=2\cdot\text{B}(2,2)\\ &=\large\color{blue}{\frac13}. \end{align}

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notas:

$\color{red}{[1]}\;\;\;u=\tan x$

$\color{red}{[2]}\;\;\;t=\sqrt{u}$

$\color{red}{[3]}\;\;\;$ Función beta :$\displaystyle\text{B}(x,y)=\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}\ dt$ #% para$\Re(x)>0$ y% #%