$\frac{π}{4} \int 0 \sqrt{1 + sin 2 x} d x =$

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Solution

$\begin{matrix} \int_{0}^{\frac{π}{4}} \sqrt{1 + sin 2 x} d x = \int_{0}^{\frac{π}{4}} \sqrt{{cos}^{2} x + {sin}^{2} x + 2 sin x cos x} d x = \int_{0}^{\frac{π}{4}} (cos x + sin x) d x = {[sin x - cos x]}_{0}^{\frac{π}{4}} = 1 \end{matrix}$

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Similar questions

\int_{0}^{π / 4} \sqrt{1 + sin 2 x} d x

\int_{0}^{π / 4} \sqrt{1 + sin 2 x} d x

\int_{0}^{\frac{π}{4}} \sqrt{\frac{1 - sin 2 x}{1 + sin 2 x}} d x =

\int_{0}^{π / 4} cos 2 x \sqrt{1 - sin 2 x} d x

I_{1} = \int_{0}^{\frac{π}{2}} \frac{{cos}^{2} x}{1 + {cos}^{2} x} d x,

I_{2} = \int_{0}^{\frac{π}{2}} \frac{{sin}^{2} x}{1 + {sin}^{2} x} d x,

I_{3} = \int_{0}^{\frac{π}{2}} \frac{1 + 2 {cos}^{2} x {sin}^{2} x}{4 + 2 {cos}^{2} x {sin}^{2} x} d x

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