Excel in math, science, and engineering

New user? Sign up

Existing user? Sign in

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

Note by Majed Musleh 1 year, 9 months ago

Sort by:

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. – Pi Han Goh · 1 year, 9 months ago

Log in to reply

@Brian Charlesworth @Pi Han Goh – Majed Musleh · 1 year, 9 months ago

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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. – Pi Han Goh · 1 year, 9 months ago

Log in to reply

@Brian Charlesworth @Pi Han Goh – Majed Musleh · 1 year, 9 months ago

Log in to reply