Explained: "Evaluate the integrals ∫0π61+sin⁡xdx"

sanuluy

Answered question

2021-10-15

Evaluate the integrals

\int_{0}^{\frac{π}{6}} \sqrt{1 + \sin x} d x

Viktor Wiley

Skilled2021-10-16Added 84 answers

Let us evaluate the given integral
Given:

\int_{0}^{\frac{π}{0}} \sqrt{1 + \sin x} d x

= \int_{0}^{\frac{π}{6}} \sqrt{1 + \sin x} \cdot \frac{\sqrt{1 - \sin x}}{\sqrt{1 - \sin x}} d x

= \int_{0}^{\frac{π}{6}} \frac{\sqrt{1 - \sin^{2} x}}{1 - \sin x} d x

= \int_{0}^{\frac{π}{6}} \frac{\sqrt{\cos^{2} x}}{\sqrt{1 - \sin x}} d x

= \int_{0}^{\frac{π}{6}} \frac{\cos x}{\sqrt{1 - \sin x}} d x

u = \sqrt{1 - \sin x}

d u = \frac{1}{2 \sqrt{1 - \sin x}} \cdot (- \cos x) d x

d u = \frac{- \cos x}{2 \sqrt{1 - \sin x}} d x

- 2 d u = \frac{\cos x}{2 \sqrt{1 - \sin x}} d x

= - 2 \int_{0}^{\frac{π}{6}}

= - 2 {[\sqrt{1 - \sin x}]}_{0}^{\frac{π}{6}}

= - 2 {[\sqrt{1 - \sin x}]}_{0}^{\frac{π}{6}}

= - 2 [\sqrt{\frac{1}{2}} - 1]

= [\frac{- 2}{\sqrt{2}} + 2]

= [\frac{2 \sqrt{2} - 2}{\sqrt{2}}]

Answer:

\int_{0}^{\frac{π}{6}} \sqrt{1 + \sin x} d x = \frac{2 \sqrt{2} - 2}{\sqrt{2}}

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