Mathematics

Evaluate $$\displaystyle\int_{0}^{\pi/4}(\sqrt{\tan}x+\sqrt{\cot}x)dx$$


SOLUTION
Let $$I=\displaystyle\int_{0}^{\pi/4}(\sqrt{\tan}x+\sqrt{\cot}x)dx$$ 

$$I=\displaystyle\int_{0}^{\pi/4}(\sqrt{\dfrac{\sin x}{\cos x}}+\sqrt{\dfrac{\cos x}{\sin x}})dx$$ 

   $$=\displaystyle\int_0^{\pi/4}\dfrac{\sin x+\cos x}{\sqrt{\sin x \cos x}}dx$$

   $$=\sqrt{2}\displaystyle\int_0^{\pi/4}\dfrac{\sin x+\cos x}{\sqrt{2\sin x \cos x}}dx$$

   $$=\sqrt{2}\displaystyle\int_0^{\pi/4}\dfrac{\sin x+\cos x}{\sqrt{1-(\sin x -\cos x)^2}}dx$$
   
Let $$\sin x-\cos x=t.$$ Then
$$\Rightarrow$$  $$\cos x+\sin x dx =dt$$            [ Differentiating both sides ]
When $$x=0,$$
$$\sin 0-\cos 0=t\Rightarrow t=1$$
When $$x=\dfrac{\pi}{4},$$
$$\sin\dfrac{\pi}{4}-\cos\dfrac{\pi}{4}=t\Rightarrow$$ t=0$$

$$\therefore$$  $$I=\sqrt{2}\displaystyle\int_{-1}^0\dfrac{dt}{\sqrt{1-t^2}}$$

        $$=\sqrt{2}\left[\sin^{-1}t\right]_{-1}^0$$
        $$=\sqrt{2}\left[\sin^{-1}(0)-\sin^{-1}(-1)\right]$$

        $$=\dfrac{\pi}{\sqrt{2}}$$
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Subjective Medium Published on 17th 09, 2020
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