$\int_0^{\frac{\pi}{2}} \frac{\cos x d x}{\sqrt{1+\cos x \sin x}}=$

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Aicharya · Accepted Answer

$\int_0^{\frac{\pi}{2}} \frac{\cos x d x}{\sqrt{1+\cos x \sin x}}$

$I=\int_0^{\frac{\pi}{2}} \frac{\cos \left(\frac{\pi}{2}-x\right)}{\sqrt{1+\cos \left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-x\right)}} d x$

On adding Eqs. (i) and (ii), we get $ \begin{aligned} 2 I & =\int_0^{\frac{\pi}{2}} \frac{\cos x+\sin x}{\sqrt{1+\cos x \sin x}} \times \frac{\sqrt{2}}{\sqrt{2}} d x \\ & =\int_0^{\frac{\pi}{2}} \frac{\sqrt{2}(\cos x+\sin x)}{\sqrt{2+2 \cos x \sin x}} \\ & =\int_0^{\frac{\pi}{2}} \frac{\sqrt{2}(\cos x+\sin x)}{\sqrt{3-(\sin x-\cos x)^2}} d x \end{aligned} $ Let $\sin x-\cos x=t$ $ (\cos x+\sin x) d x=d t $ Now, changing the limit $ \begin{aligned} & x \rightarrow 0 \Rightarrow t \rightarrow-1 \\ & x \rightarrow \pi / 2 \Rightarrow t \rightarrow 1 \\ & 2 I=\int_{-1}^1 \frac{\sqrt{2} d t}{\sqrt{3-t^2}} \\ & {\left[\int_{-a}^a f(x) d x=2 \int_0^a f(x) d x \text { if } f(x) \text { is even function }\right]} \\ & 2 I=2 \int_0^1 \frac{\sqrt{2} d t}{\sqrt{(\sqrt{3})^2-t^2}} \\ & I=\sqrt{2}\left[\sin ^{-1}\left(\frac{t}{\sqrt{3}}\right)\right]_0^1 \\ & I=\sqrt{2} \sin ^{-1}\left(\frac{1}{\sqrt{3}}\right) . \end{aligned} $

$\int_0^{\frac{\pi}{2}} \frac{\cos x d x}{\sqrt{1+\cos x \sin x}}=$

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