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# Integrate this monster

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

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.

- 2 years, 4 months ago