What is the Integral of tan^2(x)sec^2(x) dx?

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mason m Share
Jun 7, 2016

Answer:

${\tan}^{3} \frac{x}{3} + C$

Explanation:

When working with integrals of tangent and secant, it may not always be apparent what to do. Just remember that the derivative of $\tan x$ is ${\sec}^{2} x$ and the derivative of $\sec x$ is $\sec x \tan x$.

Here, notice that ${\sec}^{2} x$ is already in the integral, and all that remains is ${\tan}^{2} x$. That is, we have $\tan x$ in squared form accompanied by its derivative, ${\sec}^{2} x$. This integral is ripe for substitution!

In the integral $\int {\tan}^{2} x {\sec}^{2} x \mathrm{dx}$, let $u = \tan x$ and $\mathrm{du} = {\sec}^{2} x \mathrm{dx}$.

This gives us $\int {\tan}^{2} x {\sec}^{2} x \mathrm{dx} = \int {u}^{2} \mathrm{du}$. Performing this integration yields ${u}^{3} / 3 + C$, and since $u = \tan x$, this becomes ${\tan}^{3} \frac{x}{3} + C$.

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