Find the Integral sec(2x)tan(2x)

$\sec (2 x) \tan (2 x)$

Let $u_{2} = \sec (2 x)$ . Then $d u_{2} = 2 \sec (2 x) \tan (2 x) d x$ , so $\frac{1}{2} d u_{2} = \sec (2 x) \tan (2 x) d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Let $u_{2} = \sec (2 x)$ . Find $\frac{d u_{2}}{d x}$ .

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Differentiate $\sec (2 x)$ .

$\frac{d}{d x} [\sec (2 x)]$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sec (x)$ and $g (x) = 2 x$ .

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To apply the Chain Rule, set $u_{1}$ as $2 x$ .

$\frac{d}{d u_{1}} [\sec (u_{1})] \frac{d}{d x} [2 x]$

The derivative of $\sec (u_{1})$ with respect to $u_{1}$ is $\sec (u_{1}) \tan (u_{1})$ .

$\sec (u_{1}) \tan (u_{1}) \frac{d}{d x} [2 x]$

Replace all occurrences of $u_{1}$ with $2 x$ .

$\sec (2 x) \tan (2 x) \frac{d}{d x} [2 x]$

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$\sec (2 x) \tan (2 x) (2 \frac{d}{d x} [x])$

Move $2$ to the left of $\sec (2 x) \tan (2 x)$ .

$2 \cdot (\sec (2 x) \tan (2 x)) \frac{d}{d x} [x]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 \sec (2 x) \tan (2 x) \cdot 1$

Multiply $2$ by $1$ .

$2 \sec (2 x) \tan (2 x)$

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\int \frac{1}{2} d u_{2}$

Since $\frac{1}{2}$ is constant with respect to $u_{2}$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} u_{2} + C$

Replace all occurrences of $u_{2}$ with $\sec (2 x)$ .

$\frac{1}{2} \sec (2 x) + C$

Find the Integral sec(2x)tan(2x)

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