Question
Evaluate the following integrals: $ \int\sqrt{\cot} \theta \text{d}\theta$
✓
Answer
$ \int\sqrt{\cot}\theta\text{d}\theta$
Let $\cot\theta=\text{x}^2$
$\Rightarrow-\text{cosec}^2\theta\text{d}\theta=2\text{x dx}$
$\Rightarrow\text{d}\theta=\frac{-2\text{x}}{\text{cosec}^2\theta}\ \text{dx}$
$=\frac{-2\text{x}}{1+\cot^2\theta}$
$=\frac{-2\text{x}}{1+\text{x}^4}\ \text{dx}$
$\therefore\text{I}=-\int\frac{2\text{x}^2}{1+\text{x}^4}\ \text{dx}$
$=-\int\frac{2}{\frac{1}{\text{x}^2}+\text{x}^2}\ \text{dx}$
Dividing numerator and denominator by $x^2$
$=-\frac{1+\frac{1}{\text{x}^2}+1-\frac{1}{\text{x}^2}}{\text{x}^2+\frac{1}{\text{x}^2}}\ \text{dx}$
$=-\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}-\frac{1}{\text{x}^2}\Big)+2}-\int\frac{\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}+\frac{1}{\text{x}^2}\Big)^2-2}$
Let $\text{x}-\frac{1}{\text{x}}=\text{t}\Rightarrow\Big(1+\frac{1}{\text{x}^2}\Big)\ \text{dx}=\text{dt}$
and $\text{x}+\frac{1}{\text{x}}=\text{z}\Rightarrow\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dz}$
$\Rightarrow\text{I}=-\int\frac{\text{dt}}{\text{t}^2+2}-\int\frac{\text{dz}}{\text{z}^2-2}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{2}}\Big)-\frac{1}{2\sqrt{2}}\log\Big|\frac{\text{z}-\sqrt{2}}{\text{z}+\sqrt{2}}\Big|+\text{C}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{x}^2-1}{\sqrt{2}\text{x}}\Big)-\frac{1}{2\sqrt{2}}\log\Big|\frac{\text{x}^2+1-\sqrt{2}\text{x}}{\text{x}^2+1+\sqrt{2}\text{x}}\Big|+\text{C}$
$\text{I}=-\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\cot\theta-1}{\sqrt{2\cot\theta}}\Big)-\frac{1}{2\sqrt{2}}\log\Big|\frac{\cot\theta+1-\sqrt{2\cot\theta}}{\cot\theta+1-\sqrt{2\cot\theta}}\Big|+\text{C}$
Let $\cot\theta=\text{x}^2$
$\Rightarrow-\text{cosec}^2\theta\text{d}\theta=2\text{x dx}$
$\Rightarrow\text{d}\theta=\frac{-2\text{x}}{\text{cosec}^2\theta}\ \text{dx}$
$=\frac{-2\text{x}}{1+\cot^2\theta}$
$=\frac{-2\text{x}}{1+\text{x}^4}\ \text{dx}$
$\therefore\text{I}=-\int\frac{2\text{x}^2}{1+\text{x}^4}\ \text{dx}$
$=-\int\frac{2}{\frac{1}{\text{x}^2}+\text{x}^2}\ \text{dx}$
Dividing numerator and denominator by $x^2$
$=-\frac{1+\frac{1}{\text{x}^2}+1-\frac{1}{\text{x}^2}}{\text{x}^2+\frac{1}{\text{x}^2}}\ \text{dx}$
$=-\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}-\frac{1}{\text{x}^2}\Big)+2}-\int\frac{\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}+\frac{1}{\text{x}^2}\Big)^2-2}$
Let $\text{x}-\frac{1}{\text{x}}=\text{t}\Rightarrow\Big(1+\frac{1}{\text{x}^2}\Big)\ \text{dx}=\text{dt}$
and $\text{x}+\frac{1}{\text{x}}=\text{z}\Rightarrow\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dz}$
$\Rightarrow\text{I}=-\int\frac{\text{dt}}{\text{t}^2+2}-\int\frac{\text{dz}}{\text{z}^2-2}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{2}}\Big)-\frac{1}{2\sqrt{2}}\log\Big|\frac{\text{z}-\sqrt{2}}{\text{z}+\sqrt{2}}\Big|+\text{C}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{x}^2-1}{\sqrt{2}\text{x}}\Big)-\frac{1}{2\sqrt{2}}\log\Big|\frac{\text{x}^2+1-\sqrt{2}\text{x}}{\text{x}^2+1+\sqrt{2}\text{x}}\Big|+\text{C}$
$\text{I}=-\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\cot\theta-1}{\sqrt{2\cot\theta}}\Big)-\frac{1}{2\sqrt{2}}\log\Big|\frac{\cot\theta+1-\sqrt{2\cot\theta}}{\cot\theta+1-\sqrt{2\cot\theta}}\Big|+\text{C}$
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