## trigonometric integration question

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

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 $$\int secx{dx} = ln(tanx+secx)$$ 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)
 Recognitions: Homework Help sec(x)tan2(x)=sec3(x)sin2(x)=[sec3(x)sin(x)]sin(x), which you can integrate by parts. ehild

## trigonometric integration question

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} \int \sec^3(x)dx=\sec(x)\tan(x)-\int \sec(x)\tan^2(x)dx \\ = \sec(x)\tan(x)-\int \sec^3(x)dx + \int \sec(x)dx \\ = \sec(x)\tan(x)-\int \sec^3(x)dx + \log(\sec(x)+\tan(x)) \end{align}
Now solve that equation for the integral.