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# What is the integral of int (sinx)/(cos^2x) dx?

It is

$\int \frac{\sin x}{{\cos}^{2} x} \mathrm{dx} = \int \left[\frac{1}{\cos} x\right] ' \mathrm{dx} = \frac{1}{\cos} x + c = \sec x + c$

Mar 17, 2016

$\sec x + C$

#### Explanation:

We should try to use substitution by setting $u = \cos x$, so $\mathrm{du} = - \sin x \mathrm{dx}$.

This gives us the integral:

$\int \sin \frac{x}{\cos} ^ 2 x \mathrm{dx} = - \int \frac{- \sin x}{\cos} ^ 2 x \mathrm{dx} = - \int \frac{1}{u} ^ 2 \mathrm{du} = - \int {u}^{-} 2 \mathrm{du}$

From here, use the rule

$\int {u}^{n} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right) + C$

Thus,

$- \int {u}^{-} 2 \mathrm{du} = - {u}^{- 1} / \left(- 1\right) + C = \frac{1}{u} + C$

$= \frac{1}{\cos} x + C = \sec x + C$

Mar 17, 2016

$\sec x + C$

#### Explanation:

Alternatively, you could rewrite this in terms of other trigonometric functions:

$\int \sin \frac{x}{\cos} ^ 2 x \mathrm{dx} = \int \left(\frac{1}{\cos} x\right) \left(\sin \frac{x}{\cos} x\right) \mathrm{dx} = \int \sec x \tan x \mathrm{dx}$

If you're familiar with the fact that $\frac{d}{\mathrm{dx}} \left(\sec x\right) = \sec x \tan x$, then you'll recognize this common integral formula:

$\int \sec x \tan x \mathrm{dx} = \sec x + C$