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\[ \large \int \sqrt{\frac{\sec^{3}(x)}{1+\sin(x)}} \, dx =\ ?\]

Note by Majed Musleh 2 years, 4 months ago

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It's equals to \( \displaystyle \int \sqrt{\frac1{\cos^3(x) (1+\sin(x))}} \, dx \). Let \(y = \frac \pi2 - x \), then it becomes

\[ -\int \frac1{\sqrt{\sin^3(y)(1+\cos(y))}} \, dy \]

Apply Tangent half-angle substitution, then it equals to

\[ - \int \sqrt{\frac{1}{\left(\frac{2t}{1+t^2}\right)^3\left(1+ \frac{1-t^2}{1+t^2}\right)} }\cdot \frac{2 dt}{t^2+1} = -\frac12 \int \frac{1+t^2}{t^{3/2}} \, dt \]

Which can be easily integrated from here, back substitute everything and you're done.

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@Brian Charlesworth @Pi Han Goh

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TopNewestIt's equals to \( \displaystyle \int \sqrt{\frac1{\cos^3(x) (1+\sin(x))}} \, dx \). Let \(y = \frac \pi2 - x \), then it becomes

\[ -\int \frac1{\sqrt{\sin^3(y)(1+\cos(y))}} \, dy \]

Apply Tangent half-angle substitution, then it equals to

\[ - \int \sqrt{\frac{1}{\left(\frac{2t}{1+t^2}\right)^3\left(1+ \frac{1-t^2}{1+t^2}\right)} }\cdot \frac{2 dt}{t^2+1} = -\frac12 \int \frac{1+t^2}{t^{3/2}} \, dt \]

Which can be easily integrated from here, back substitute everything and you're done.

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@Brian Charlesworth @Pi Han Goh

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