Evaluate the following integral: \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx

Michael Maggard 2022-01-06 Answered
Evaluate the following integral:
\(\displaystyle\int{\frac{{\sqrt{{{\sin{{x}}}}}}}{{\sqrt{{{\sin{{x}}}}}+\sqrt{{{\cos{{x}}}}}}}}{\left.{d}{x}\right.}\)

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Expert Answer

Fasaniu
Answered 2022-01-07 Author has 677 answers
\(\displaystyle{I}=\int{\frac{{\sqrt{{{\sin{{x}}}}}}}{{\sqrt{{{\sin{{x}}}}}+\sqrt{{{\cos{{x}}}}}}}}{\left.{d}{x}\right.}=\int{\frac{{\sqrt{{{\tan{{x}}}}}}}{{{1}+\sqrt{{{\tan{{x}}}}}}}}{\left.{d}{x}\right.}\)
substitute \(\displaystyle{\tan{{x}}}={t}^{{2}}\) and \(\displaystyle{\left.{d}{x}\right.}={\frac{{{1}}}{{{1}+{t}^{{4}}}}}{\left.{d}{t}\right.}\)
\(\displaystyle{I}=\int{\frac{{{t}}}{{{\left({1}+{t}\right)}{\left({1}+{t}^{{4}}\right)}}}}{\left.{d}{t}\right.}={\frac{{{1}}}{{{2}}}}\int{\frac{{{\left({\left({1}+{t}^{{4}}\right)}+{\left({1}-{t}^{{4}}\right)}\right)}{t}}}{{{\left({1}+{t}\right)}{\left({1}+{t}^{{4}}\right)}}}}{\left.{d}{t}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}\int{\frac{{{t}}}{{{1}+{t}}}}{\left.{d}{t}\right.}+{\frac{{{1}}}{{{2}}}}\int{\frac{{{\left({t}-{t}^{{2}}\right)}{\left({1}+{t}^{{2}}\right)}}}{{{1}+{t}^{{4}}}}}{\left.{d}{t}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}\int{\frac{{{t}}}{{{1}+{t}}}}{\left.{d}{t}\right.}+{\frac{{{1}}}{{{2}}}}\int{\frac{{{\left({t}-{t}^{{2}}\right)}{\left({1}+{t}^{{2}}\right)}}}{{{1}+{t}^{{4}}}}}{\left.{d}{t}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}\int{\frac{{{\left({1}+{t}\right)}-{1}}}{{{1}+{t}}}}{\left.{d}{t}\right.}+{\frac{{{1}}}{{{2}}}}\int{\frac{{{t}+{t}^{{3}}-{\left({t}^{{2}}-{1}\right)}-{t}^{{4}}-{1}}}{{{1}+{t}^{{4}}}}}{\left.{d}{t}\right.}\)
\(\displaystyle=-{\frac{{{t}}}{{{2}}}}+{\frac{{{1}}}{{{2}}}}{\ln}{\left|{t}+{1}\right|}+{\frac{{{1}}}{{{4}}}}\int{\frac{{{2}{t}}}{{{1}+{t}^{{4}}}}}+{\frac{{{1}}}{{{2}}}}\int{\frac{{{t}^{{3}}}}{{{1}+{t}^{{4}}}}}{\left.{d}{t}\right.}-{\frac{{{1}}}{{{2}}}}\int{\frac{{{t}^{{2}}-{1}}}{{{1}+{t}^{{4}}}}}{\left.{d}{t}\right.}-{\frac{{{1}}}{{{2}}}}+{C}\)
All integrals are easy expect
\(\displaystyle{I}=\int{\frac{{{t}^{{2}}-{1}}}{{{1}+{t}^{{4}}}}}{\left.{d}{t}\right.}=\int{\frac{{{1}-{t}^{{-{2}}}}}{{{\left({t}+{t}^{{-{1}}}\right)}^{{2}}-{2}}}}{\left.{d}{t}\right.}=\int{\frac{{{\left({t}-{t}^{{-{1}}}\right)}'}}{{{\left({t}-{t}^{{-{1}}}\right)}^{{2}}-{2}}}}{\left.{d}{t}\right.}\)
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usaho4w
Answered 2022-01-08 Author has 2871 answers
We rationalise the denominator to get
\(\displaystyle{I}=\int{\frac{{{\sin{{x}}}-\sqrt{{{\cos{{x}}}{\sin{{x}}}}}}}{{{\sin{{x}}}-{\cos{{x}}}}}}{\left.{d}{x}\right.}\)
Writing everything in terms of \(\displaystyle{\cot{{x}}}\), we get
\(\displaystyle{I}=\int{{\csc}^{{2}}{x}}{\left({\frac{{\sqrt{{{\cot{{x}}}}}-{1}}}{{{{\cot}^{{3}}{x}}-{{\cot}^{{2}}{x}}+{\cot{{x}}}-{1}}}}\right)}{\left.{d}{x}\right.}\)
Now substituting \(\displaystyle{u}={\cot{{x}}}\) and further \(\displaystyle{v}=\sqrt{{{u}}}\) gives us
\(\displaystyle{I}=-\int{\frac{{{2}{v}}}{{{v}^{{5}}+{v}^{{4}}+{v}+{1}}}}{d}{v}\)
Performing a partial fraction decomposition we have
\(\displaystyle{I}={\frac{{{2}}}{{{4}-{4}\sqrt{{{2}}}}}}\int{\frac{{{v}+\sqrt{{{2}}}-{1}}}{{{v}^{{2}}+\sqrt{{{2}}}{v}+{1}}}}{d}{v}+{\frac{{{2}}}{{{4}+{4}\sqrt{{{2}}}}}}\)
\(\displaystyle\int{\frac{{{v}-\sqrt{{{2}}}-{1}}}{{{v}^{{2}}-\sqrt{{{2}}}{v}+{1}}}}{d}{v}+\int{\frac{{{1}}}{{{1}+{v}}}}{d}{v}={I}_{{1}}+{I}_{{2}}+{I}_{{3}}\)
Hope you can take it from here
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star233
Answered 2022-01-11 Author has 0 answers

HINT multiply nominator and denominator by \(\frac{1}{\sqrt{\sin(x)}}\), then \(t=\sqrt{\cot(x)}\) after all you'll have \(\frac{2t}{(t^4+1)(t+1)}\)

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