Integrals of Trigonometric Functions

Integrating Even Powers of Sine and Cosine

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 2 videos by turksvids

Key Questions

• Since

$\left(\sin x\right) ' = \cos x$,

we have

$\int \cos x \mathrm{dx} = \sin x + C$.

Since

$\left(- \cos x\right) ' = \sin x$,

we have

$\int \sin x \mathrm{dx} = - \cos x + C$.

I hope that this was helpful.

A different approach...

Explanation:

We want to find $\int \tan x \mathrm{dx}$.

$\int \tan x \mathrm{dx} = \int \frac{\tan x \sec x}{\sec} x \mathrm{dx}$

Now let $u = \sec x$ and $\mathrm{du} = \sec x \tan x \mathrm{dx}$. Then

$\int \frac{\tan x \sec x}{\sec} x \mathrm{dx} = \int \frac{1}{u} \mathrm{du}$

This is a standard integral which evaluates to

$\ln \left\mid u \right\mid + \text{c"=lnabssecx+"c}$

• Since

$\left(\tan x\right) ' = {\sec}^{2} x$,

we have

$\int {\sec}^{2} x \mathrm{dx} = \tan x + C$.

I hope that this was helpful.

• Since

$\left(\ln | \sec x + \tan x |\right) ' = \frac{\sec x \tan x + {\sec}^{2} x}{\sec x + \tan x} = \sec x$,

we have

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

Since

$\left(- \ln | \csc x + \cot x |\right) ' = - \frac{- \csc x \cot x - {\csc}^{2} x}{\csc x + \cot x} = \csc x$,

we have

$\int \csc x \mathrm{dx} = - \ln | \csc x + \cot x | + C$

$\int \cot x \mathrm{dx} = \int \frac{\cos x}{\sin x} \mathrm{dx} = \ln | \sin x | + C$

I hope that this was helpful.

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