What is the integral of 1/((sec(x)+tan(x))^2) ?

The integral of 1/((sec(x)+tan(x))^2) is 2(-2/(tan(x/2)+1)-arctan(tan(x/2)))+C

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integral of 1/((sec(x)+tan(x))^2)

\int \frac{1}{( sec ( x ) + tan ( x ) ) ^{2}} d x

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

Solution steps

\int \frac{1}{( sec ( x ) + tan ( x ) ) ^{2}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Take the partial fraction of

\frac{( u - 1 ) ^{2}}{( u + 1 ) ^{2} ( 1 + u ^{2} )} : \frac{2}{( u + 1 ) ^{2}} - \frac{1}{u ^{2} + 1}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= 2 (- \frac{2}{u + 1} - arctan (u))

Substitute back

u = tan (\frac{x}{2})

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (3 x e^{- \frac{1}{y ^{3}}})

\int_{115}^{150} \frac{75}{2 x ^{2}} d x

\frac{( x ^{2} - 3 x )}{x ^{2} + 7 x}

y = 9 - x

\frac{d}{d x} (x (4 - x^{2})^{2})

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