What is the integral of (tan(x))/(cos(2x)) ?

The integral of (tan(x))/(cos(2x)) is -1/2 ln|1-tan^2(x)|+C

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integral of (tan(x))/(cos(2x))

\int \frac{tan ( x )}{cos ( 2 x )} d x

- \frac{1}{2} ln 1 - tan^{2} (x) + C

Solution steps

\int \frac{tan ( x )}{cos ( 2 x )} d x

Apply u-substitution

= \int \frac{tan ( \frac{u}{2} )}{2 cos ( u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{tan ( \frac{u}{2} )}{cos ( u )} d u

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{2 v}{1 - v ^{2}} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot 2 \cdot \int \frac{v}{1 - v ^{2}} d v

Apply u-substitution

= \frac{1}{2} \cdot 2 \cdot \int - \frac{1}{2 w} d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot 2 (- \frac{1}{2} \cdot \int \frac{1}{w} d w)

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= \frac{1}{2} \cdot 2 (- \frac{1}{2} ln ∣ w ∣)

Substitute back

= \frac{1}{2} \cdot 2 (- \frac{1}{2} ln 1 - tan^{2} (\frac{2 x}{2}))

Simplify

\frac{1}{2} \cdot 2 (- \frac{1}{2} ln 1 - tan^{2} (\frac{2 x}{2})) : - \frac{1}{2} ln 1 - tan^{2} (x)

= - \frac{1}{2} ln 1 - tan^{2} (x)

Add a constant to the solution

= - \frac{1}{2} ln 1 - tan^{2} (x) + C

What is the integral of (tan(x))/(cos(2x)) ?
The integral of (tan(x))/(cos(2x)) is -1/2 ln|1-tan^2(x)|+C