Evaluate the following integrals: _ 0 ^ 4 ( + )dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{4}}\Big(\sqrt{\tan\text{x}}+\sqrt{\cot}\text{x}\Big)\text{dx}$

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Answer

Let $\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\Big(\sqrt{\tan\text{x}}+\sqrt{\cot}\text{x}\Big)\text{dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\bigg(\sqrt{\frac{\sin\text{x}}{\cos\text{x}}}+\sqrt{\frac{\cos\text{x}}{\sin\text{x}}}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{\sin\text{x}\cos\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\sqrt{2}\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2\sin\text{x}\cos\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\sqrt{2}\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1-(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
Let $\sin\text{x}-\cos\text{x}=\text{t}$ Then, $\cos\text{x}+\sin\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=1$ and $\text{x}=\frac{\pi}{4},\text{t}=0$
$\therefore\ \text{I}=\sqrt{2}\int^\limits0_{-1}\frac{\text{dt}}{\sqrt{1-\text{t}^2}}$
$\Rightarrow\text{I}=\sqrt{2}\Big[\sin^{-1}\text{t}\Big]^0_{-1}$
$\Rightarrow\text{I}=\frac{\pi}{\sqrt{2}}$

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