Trigonometric integration question

[stonecoldgen]

The question asks to find ∫secxtan²x

I rewrote tan²x as (sec²x-1). Then I expanded the equation having sec³x-secx and I know the integral of secx which is 0.5ln|tanx+secx|,

but my question is, is integrating sec³x by parts the correct path? or not?

Thanks

 

[AGNuke]

\int secx{dx} = ln(tanx+secx)

And as for ∫sec³x dx; You can try parts, but you thought how you can break it?
OR you can look for some formula of finding integrals of powers of trigonometric functions (Reduction formulas)

 

[ehild]

sec(x)tan²(x)=sec³(x)sin²(x)=[sec³(x)sin(x)]sin(x), which you can integrate by parts.

ehild

 

[Millennial]

The integral of secant cubed can be evaluated as follows (it is a common integral) with using integration by parts, applying u=\sec(x) and dv=\sec^2(x)dx:
\begin{align}<br /> \int \sec^3(x)dx=\sec(x)\tan(x)-\int \sec(x)\tan^2(x)dx \\<br /> = \sec(x)\tan(x)-\int \sec^3(x)dx + \int \sec(x)dx \\<br /> = \sec(x)\tan(x)-\int \sec^3(x)dx + \log(\sec(x)+\tan(x))<br /> \end{align}
Now solve that equation for the integral.