Evaluate the integrals \int_0^{\pi/6}\sqrt{1+\sin x}dx

sanuluy 2021-10-15 Answered
Evaluate the integrals
\(\displaystyle{\int_{{0}}^{{\frac{\pi}{{6}}}}}\sqrt{{{1}+{\sin{{x}}}}}{\left.{d}{x}\right.}\)

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Expert Answer

Viktor Wiley
Answered 2021-10-16 Author has 17756 answers
Let us evaluate the given integral
Given: \(\displaystyle{\int_{{0}}^{{\frac{\pi}{{0}}}}}\sqrt{{{1}+{\sin{{x}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{6}}}}}\sqrt{{{1}+{\sin{{x}}}}}\cdot{\frac{{\sqrt{{{1}-{\sin{{x}}}}}}}{{\sqrt{{{1}-{\sin{{x}}}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{6}}}}}{\frac{{\sqrt{{{1}-{{\sin}^{{2}}{x}}}}}}{{{1}-{\sin{{x}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{6}}}}}{\frac{{\sqrt{{{{\cos}^{{2}}{x}}}}}}{{\sqrt{{{1}-{\sin{{x}}}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{6}}}}}{\frac{{{\cos{{x}}}}}{{\sqrt{{{1}-{\sin{{x}}}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle{u}=\sqrt{{{1}-{\sin{{x}}}}}\)
\(\displaystyle{d}{u}={\frac{{{1}}}{{{2}\sqrt{{{1}-{\sin{{x}}}}}}}}\cdot{\left(-{\cos{{x}}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle{d}{u}={\frac{{-{\cos{{x}}}}}{{{2}\sqrt{{{1}-{\sin{{x}}}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle-{2}{d}{u}={\frac{{{\cos{{x}}}}}{{{2}\sqrt{{{1}-{\sin{{x}}}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle=-{2}{\int_{{0}}^{{\frac{\pi}{{6}}}}}\)
\(\displaystyle=-{2}{{\left[\sqrt{{{1}-{\sin{{x}}}}}\right]}_{{0}}^{{\frac{\pi}{{6}}}}}\)
\(\displaystyle=-{2}{{\left[\sqrt{{{1}-{\sin{{x}}}}}\right]}_{{0}}^{{\frac{\pi}{{6}}}}}\)
\(\displaystyle=-{2}{\left[\sqrt{{{\frac{{{1}}}{{{2}}}}}}-{1}\right]}\)
\(\displaystyle={\left[{\frac{{-{2}}}{{\sqrt{{{2}}}}}}+{2}\right]}\)
\(\displaystyle={\left[{\frac{{{2}\sqrt{{{2}}}-{2}}}{{\sqrt{{{2}}}}}}\right]}\)
Answer: \(\displaystyle{\int_{{0}}^{{\frac{\pi}{{6}}}}}\sqrt{{{1}+{\sin{{x}}}}}{\left.{d}{x}\right.}={\frac{{{2}\sqrt{{{2}}}-{2}}}{{\sqrt{{{2}}}}}}\)
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