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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{1}{1 + tan ( x )} d x

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

\int \frac{1}{2 ( u + 1 )} d u = \frac{1}{2} ln ∣ u + 1 ∣

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

Substitute back

u = tan (x)

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

Simplify

2 (\frac{1}{2} (- \frac{1}{2} ln tan^{2} (x) + 1 + arctan (tan (x))) + \frac{1}{2} ln ∣ tan (x) + 1 ∣) : - \frac{1}{2} ln (sec^{2} (x)) + arctan (tan (x)) + ln ∣ tan (x) + 1 ∣

= - \frac{1}{2} ln (sec^{2} (x)) + arctan (tan (x)) + ln ∣ tan (x) + 1 ∣

Add a constant to the solution

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

\int_{4}^{8} t^{4} ln (3 t) d t

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\frac{d}{d x} (e^{2 x^{2} - 5 x})

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

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