# What is the integral of #int tan^3(x) dx#?

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This is a common integral:

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The integral of ( \int \tan^3(x) , dx ) can be evaluated using the substitution method. Let ( u = \tan(x) ), then ( du = \sec^2(x) , dx ). This implies ( dx = \frac{du}{\sec^2(x)} = \cos^2(x) , du ).

So the integral becomes:

[ \int \tan^3(x) , dx = \int u^3 \cos^2(x) , du ]

Now, we need to express ( \cos^2(x) ) in terms of ( u ). Since ( u = \tan(x) ), we have ( \tan^2(x) + 1 = \sec^2(x) ). This implies ( 1 + u^2 = \sec^2(x) ), so ( \cos^2(x) = \frac{1}{1 + u^2} ).

Substituting this into the integral:

[ \int \tan^3(x) , dx = \int u^3 \cdot \frac{1}{1 + u^2} , du ]

This is a rational function and can be integrated by techniques like partial fraction decomposition. After integration, you'll get the final result.

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