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

[tex]\int secx{dx} = ln(tanx+secx)[/tex]

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 [itex]u=\sec(x)[/itex] and [itex]dv=\sec^2(x)dx[/itex]:
[tex]\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}[/tex]
Now solve that equation for the integral.