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Indefinite integral of secant cubed

I guess I need to learn a new technique because those I know didn't help me here. Wolframalpha uses the reduction formula, which I haven't been introduced to yet. So is there a way to solve this without using the reduction formula?

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marked as duplicate by Hans Lundmark, Noah Snyder, Martin Sleziak, Hagen von Eitzen, Emily Oct 17 '12 at 18:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

...and this one too:… – Hans Lundmark Oct 17 '12 at 9:47
up vote 2 down vote accepted

I use integration by parts and $\tan^2\theta = \sec^2\theta - 1$. \begin{align*} \int\sec^3\theta\ d\theta & = \int\sec\theta\ d(\tan\theta) \\ & = \sec\theta\tan\theta - \int \tan\theta\ d(\sec\theta) \\ & = \sec\theta\tan\theta - \int \sec\theta\tan^2\theta\ d\theta \\ & = \sec\theta\tan\theta - \int \sec\theta(\sec^2\theta - 1)\ d\theta \\ & = \sec\theta\tan\theta + \int\sec\theta\ d\theta - \int \sec^3\theta\ d\theta \\ \therefore 2 \int\sec^3\theta\ d\theta & = \sec\theta\tan\theta + \int\sec\theta\ d\theta. \end{align*}

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It is complete for the OP. +1 – Babak S. Nov 27 '12 at 20:33

Yes; it involves a couple of integrations by parts and some manipulation with trig identities.

Let $I=\int\sec^3\theta~d\theta$. Let $u=\sec\theta$ and $dv=\sec^2\theta~d\theta$; then $du=\sec\theta\tan\theta~d\theta$ and $v=\tan\theta$, so


Now $\tan^2\theta=\sec^2\theta-1$, so


Combining $(1)$ and $(2)$ yields

$$I=\sec\theta\tan\theta-\left(I-\int\sec\theta~d\theta\right)=\sec\theta\tan\theta-I+\int\sec\theta~d\theta\;,$$ so


But $$\int\sec\theta~d\theta=\int\frac{\sec\theta(\sec\theta+\tan\theta)}{\sec\theta+\tan\theta}d\theta=\int\frac{d(\sec\theta+\tan\theta)}{\sec\theta+\tan\theta}=\ln(\sec\theta+\tan\theta)+C\;,$$ so finally


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Make the change of variables $ \sec(x)^2 = 1-u $, which implies

$$ \int \sec(\theta)^3 d \theta = -\frac{1}{2}\int \frac{\sqrt{u-1}}{\sqrt{u}} du \,. $$

Following it with the change of variables $ u = \sin(t)^2 $ yields

$$ -\frac{1}{2}\int \frac{\sqrt{u-1}}{\sqrt{u}} du = -i \int \cos(t)^2\,dz= -i \left(\frac{1}{4} \sin \left( 2 t \right) + \frac{t}{2} \right)\,. $$

Substituting $ t = \arcsin{\sqrt{u}} $ and then $ u = -\tan(x)^2 $ gives the desired result.

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