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  4. The value of the definite integral ∫ limits0(π/6) (√1+ sin x/ cos x) d x is equal to

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Q. The value of the definite integral $\int\limits_0^{\frac{\pi}{6}} \frac{\sqrt{1+\sin x}}{\cos x} d x$ is equal to

Integrals

Solution:

$I=\int\limits_0^{\frac{\pi}{6}} \frac{\sqrt{1+\sin x}}{\cos x} d x=\int\limits_0^{\frac{\pi}{6}} \frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}} d x=\int\limits_0^{\frac{\pi}{6}} \frac{d x}{\cos \frac{x}{2}-\sin \frac{x}{2}}=\int\limits_0^{\frac{\pi}{6}} \frac{d x}{\sqrt{2} \cos \left(\frac{x}{2}+\frac{\pi}{4}\right)}$
$=\frac{1}{\sqrt{2}} \int\limits_0^{\frac{\pi}{6}} \sec \left(\frac{x}{2}+\frac{\pi}{4}\right) dx =\sqrt{2}\left[\ln \left(\sec \left(\frac{\pi}{4}+\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right)\right]_0^{\frac{\pi}{6}}$
$=\sqrt{2} \ln \left(\frac{2+\sqrt{3}}{\sqrt{2}+1}\right)=\sqrt{2} \ln (\sqrt{8}+\sqrt{6}-\sqrt{4}-\sqrt{3})$