Calculus Examples

$\int \frac{\tan (x)}{\ln (\cos (x))} d x$

Step 1

Let $u_{2} = \ln (\cos (x))$ . Then $d u_{2} = - \tan (x) d x$ , so $- d u_{2} = \tan (x) d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 1.1

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

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Step 1.1.1

Differentiate $\ln (\cos (x))$ .

$\frac{d}{d x} [\ln (\cos (x))]$

Step 1.1.2

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

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Step 1.1.2.1

To apply the Chain Rule, set $u_{1}$ as $\cos (x)$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [\cos (x)]$

Step 1.1.2.2

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

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

Step 1.1.2.3

Replace all occurrences of $u_{1}$ with $\cos (x)$ .

$\frac{1}{\cos (x)} \frac{d}{d x} [\cos (x)]$

Step 1.1.3

Convert from $\frac{1}{\cos (x)}$ to $\sec (x)$ .

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

Step 1.1.4

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$\sec (x) (- \sin (x))$

Step 1.1.5

Simplify.

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Step 1.1.5.1

Reorder the factors of $\sec (x) (- \sin (x))$ .

$- \sec (x) \sin (x)$

Step 1.1.5.2

Rewrite $\sec (x)$ in terms of sines and cosines.

$- \frac{1}{\cos (x)} \sin (x)$

Step 1.1.5.3

Combine $\sin (x)$ and $\frac{1}{\cos (x)}$ .

$- \frac{\sin (x)}{\cos (x)}$

Step 1.1.5.4

Convert from $\frac{\sin (x)}{\cos (x)}$ to $\tan (x)$ .

$- \tan (x)$

Step 1.2

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

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

Step 2

Split the fraction into multiple fractions.

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

Step 3

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

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

Step 4

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$- (\ln (| u_{2} |) + C)$

Step 5

Simplify.

$- \ln (| u_{2} |) + C$

Step 6

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

$- \ln (| \ln (\cos (x)) |) + C$