# How do I solve $\int \sec^3 \theta d\theta$ [duplicate]

Possible Duplicate:
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, ArkamisOct 17 '12 at 18:44

...and this one too: math.stackexchange.com/questions/179935/… –  Hans Lundmark Oct 17 '12 at 9:47

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

$$I=\sec\theta\tan\theta-\int\sec\theta\tan^2\theta~d\theta\;.\tag{1}$$

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

$$\int\sec\theta\tan^2\theta~d\theta=\int\left(\sec^3\theta-\sec\theta\right)d\theta=I-\int\sec\theta~d\theta\;.\tag{2}$$

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

$$I=\frac12\left(\sec\theta\tan\theta+\int\sec\theta~d\theta\right)\;.$$

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

$$I=\frac12\sec\theta\tan\theta+\frac12\ln(\sec\theta+\tan\theta)+C\;.$$

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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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