What is the integral of sec(x)[sec(x)-tan(x)] ?

The integral of sec(x)[sec(x)-tan(x)] is -2/(tan(x/2)+1)+C

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integral of sec(x)[sec(x)-tan(x)]

\int sec (x) [sec (x) - tan (x)] d x

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

Solution steps

\int sec (x) [sec (x) - tan (x)] d x

Simplify

sec (x) [sec (x) - tan (x)] : \frac{1}{1 + sin ( x )}

= \int \frac{1}{1 + sin ( x )} d x

Apply u-substitution

: \int \frac{2}{1 + u ^{2} + 2 u} d u

= \int \frac{2}{1 + u ^{2} + 2 u} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{1 + u ^{2} + 2 u} d u

Complete the square

1 + u^{2} + 2 u : (u + 1)^{2}

= 2 \cdot \int \frac{1}{( u + 1 ) ^{2}} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

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

Apply the Power Rule

: - \frac{1}{v}

= 2 (- \frac{1}{v})

Substitute back

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

Simplify

2 (- \frac{1}{tan ( \frac{x}{2} ) + 1}) : - \frac{2}{tan ( \frac{x}{2} ) + 1}

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

Add a constant to the solution

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

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What is the integral of sec(x)[sec(x)-tan(x)] ?
The integral of sec(x)[sec(x)-tan(x)] is -2/(tan(x/2)+1)+C