# What is the integral of #int cos^2(x) tan^3(x) dx#?

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The integral of ( \int \cos^2(x) \tan^3(x) , dx ) can be solved using trigonometric identities. By applying the identity ( \tan^2(x) = \sec^2(x) - 1 ), and then expressing ( \cos^2(x) ) in terms of ( \sec^2(x) ), the integral can be rewritten in a form that can be more easily integrated.

After applying the trigonometric identities, the integral simplifies to ( \int (\sec^2(x) - 1) \tan^3(x) , dx ).

Now, let ( u = \tan(x) ), then ( du = \sec^2(x) , dx ).

Substituting these into the integral, we get ( \int (u^3 - u) , du ).

Integrating term by term, we obtain ( \frac{1}{4}u^4 - \frac{1}{2}u^2 + C ), where ( C ) is the constant of integration.

Finally, substituting back ( u = \tan(x) ), the integral becomes ( \frac{1}{4}\tan^4(x) - \frac{1}{2}\tan^2(x) + C ), where ( C ) is the constant of integration.

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