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I have to solve for the integral: $$\int \sec^3 x \ dx$$ I used integration by parts, letting: $$u=\sec x$$ $$du = \sec x \tan x \ dx$$ $$dv=\sec^2 x$$ $$v=\tan x$$ Integration by parts formula: $$\int u \ dv = uv - \int v \ du$$ Using integration by parts: $$\sec x \tan x - \int \tan x \sec x \tan x \ dx$$ $$\sec x \tan x - \int \tan^2 x \sec x \ dx$$ Using the identity $\tan^2 x = \sec^2 x -1$: $$\sec x \tan x - \int \sec x \left(\sec^2 x - 1\right) \ dx$$ $$\sec x \tan x - \int \sec^3 x \ dx + \int \sec x \ dx$$ $$\sec x \tan x + \ln|\sec x \tan x| - \int \sec^3 x \ dx$$ What do I do now? If I integrate $\int \sec^3 x \ dx$ again, I will just continue going in an infinite loop and end up with $\int \sec^3 x \ dx$ again. Can someone please point me in the right direction? Thanks

Anonymous Computer
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    Add the integral you're solving for to both sides of the equation and then divide by 2. – Nate Chandler Mar 10 '14 at 00:48
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    The reappearance of some multiple (including 1) of the original integral appearing on the "right-hand side" is the basis for a great many "reduction formulas" for indefinite integrals as well. – colormegone Mar 11 '14 at 07:41

3 Answers3

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Read the Wikipedia page at http://en.wikipedia.org/wiki/Integral_of_secant_cubed - this should give you your answer.

You could continue with what you've done and use the integration constant $C$ to get $$\int\sec^3(x)dx=\dfrac{1}{2}\left(C+\sec(x)\tan(x)+\ln|\sec(x)\tan(x)|\right)$$

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If you did your algebra correctly, you're actually done. Let $I=\int\sec^3(x)\,dx$. Then you've demonstrated $I=f(x)-I \implies I = f(x)/2$, where $f(x)$ is all that $\sec{x}\tan{x}$ business you have.

Jason
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It may help to write down the whole equation you have shown:

$$ \int \sec^3 x \, dx = C + \sec x \tan x + \ln|\sec x \tan x| - \int \sec^3 x \ dx $$

Now, what do you usually do when the thing you want to solve for appears multiple times in an equation? :)

(I've added the "+C" because otherwise, the two copies of $\int \sec^3 x \, dx$ might differ by a constant. Adding in the "+C" allows us to ensure they are actually the same antiderivative of $\sec^3 x$)