Evaluate the following integrals: ^ 2 _ - 2 - 2 ^2x dx

Question

Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{-\frac{\pi}{2}}{\sqrt{\cos\text{x}\sin^2\text{x}}}\text{ dx}$

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Answer

Let $\text{I}=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{-\frac{\pi}{2}}{\sqrt{\cos\text{x}\sin^2\text{x}}}\text{ dx}$$=-\frac{\pi}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{1}{\sqrt{\cos\text{x}\sin^2\text{x}}}\text{ dx}$
$=-\frac{\pi}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{1}{\sqrt{\cos\text{x}}\ |\sin\text{x}|}\text{ dx}$
$=-\frac{\pi}{2}\times2\int\limits^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\cos\text{x}}\ |\sin\text{x}|}\text{ dx}$ $\Big[\text{f}(-\text{x})=\sqrt{\cos(-\text{x})}\ |\sin(-\text{x})|=\sqrt{\cos\text{x}}\ |-\sin\text{x}|=\sqrt{\cos\text{x}}\ |\sin\text{x}|=\text{f(x)}\Big]$
$=-\pi\int\limits^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\cos\text{x }}\sin\text{x}}\text{ dx}$ $\Big(|\sin\text{x}|=\sin\text{x},0\leq\text{x}\leq\frac{\pi}{2}\Big)$
$=-\pi\int\limits^{\frac{\pi}{2}}_{0}\frac{\sin\text{x}}{\sqrt{\cos\text{x }}(1-\cos^2\text{x})}\text{ dx}$
Put $\cos\text{x}=\text{z}^2$$\therefore\ -\sin\text{x dx}=2\text{zdz}$
When $\text{x}\rightarrow0,\text{z}\rightarrow1$ When $\text{x}\rightarrow\frac{\pi}{2},\text{z}\rightarrow0$$\therefore\ \text{I}=2\pi\int\limits^0_1\frac{\text{zdz}}{\text{z}(1-\text{z}^4)}$
$=2\pi\int\limits^0_1\frac{\text{dz}}{(1-\text{z}^4)}$
$=2\pi\int\limits^0_1\frac{\text{dz}}{(1-\text{z})(1+\text{z})(1+\text{z}^2)}$
Now,$\frac{1}{(1-\text{z})(1+\text{z})(1+\text{z}^2)}=\frac{\text{A}}{1-\text{z}}+\frac{\text{B}}{1+\text{z}}+\frac{\text{Cz}+\text{D}}{1+\text{z}^2}$
$1=\text{A}(1+\text{z})(1+\text{z}^2)+\text{B}(1-\text{z})(1+\text{z}^2)+(\text{Cz}+\text{D})(1-\text{z})(1+\text{z})$
Putting z = 1, we get$\text{A}=\frac{1}{4}$
Putting z = -1$\text{B}=\frac{1}{4}$
Putting z = 0$1=\text{A}+\text{B}+\text{D}$
$\text{D}=1-\frac{1}{4}-\frac{1}{4}=\frac{1}{2}$
$\text{D}=\frac{1}{2}$
Equating coefficient of $z^3$on both side, we get$\text{A}-\text{B}+\text{C}=0$
$\frac{1}{4}-\frac{1}{4}+\text{C}=0$
$\text{C}=0$
$\therefore\ \text{I}=2\pi\int\limits^0_1\frac{\text{dz}}{(1-\text{z})(1+\text{z})(1+\text{z}^2)}$
$=2\pi\int\limits^0_1\frac{\frac{1}{4}}{1-\text{z}}\text{dz}+2\pi\int\limits^0_1\frac{\frac{1}{4}}{1+\text{z}}\text{dz}+2\pi\int\limits^0_1\frac{\frac{1}{2}}{1+\text{z}^2}\text{dz}$
$=\frac{2\pi}{4}\times\bigg[\frac{\log(1-\text{z})}{-1}\bigg]^0_1+\frac{2\pi}{4}\times\big[\log(1+\text{z})\big]^0_1+\frac{2\pi}{2}\times\big[\tan^{-1}\text{z}\big]^0_1$
$=-\frac{\pi}{2}\big(\log1-\log0\big)+\frac{\pi}{2}\big(\log1-\log2\big)+\pi\big(\tan^{-1}0-\tan^{-1}1\big)$
$=-\frac{\pi}{2}\big[0-(-\infty)\big]+\frac{\pi}{2}(0-\log2)+\pi\Big(0-\frac{\pi}{4}\Big)$
$=-\infty-\frac{\pi}{2}\log2-\frac{\pi^2}{4}$
$=-\infty$

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