What is the integral of 1/(1-tan(6x)) ?

The integral of 1/(1-tan(6x)) is 1/12 (-ln|-1+tan^2(3x)+2tan(3x)|+ln(sec^2(3x))+2arctan(tan(3x)))+C

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integral of 1/(1-tan(6x))

\int \frac{1}{1 - tan ( 6 x )} d x

\frac{1}{12} (- ln - 1 + tan^{2} (3 x) + 2 tan (3 x) + ln (sec^{2} (3 x)) + 2 arctan (tan (3 x))) + C

Solution steps

\int \frac{1}{1 - tan ( 6 x )} d x

Apply u-substitution

: \int \frac{1}{6 ( 1 - tan ( u ) )} d u

= \int \frac{1}{6 ( 1 - tan ( u ) )} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{1 - tan ( u )} d u

Apply u-substitution

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

= \frac{1}{6} \cdot \int \frac{2 ( 1 - v ^{2} )}{( 1 - v ^{2} - 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}{6} \cdot 2 \cdot \int \frac{1 - v ^{2}}{( 1 - v ^{2} - 2 v ) ( 1 + v ^{2} )} d v

Take the partial fraction of

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

= \frac{1}{6} \cdot 2 \cdot \int \frac{- v - 1}{2 ( v ^{2} + 2 v - 1 )} + \frac{v + 1}{2 ( v ^{2} + 1 )} d v

Apply the Sum Rule:

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

= \frac{1}{6} \cdot 2 (\int \frac{- v - 1}{2 ( v ^{2} + 2 v - 1 )} d v + \int \frac{v + 1}{2 ( v ^{2} + 1 )} d v)

\int \frac{- v - 1}{2 ( v ^{2} + 2 v - 1 )} d v = - \frac{1}{4} ln v^{2} + 2 v - 1

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

= \frac{1}{6} \cdot 2 (- \frac{1}{4} ln v^{2} + 2 v - 1 + \frac{1}{2} (\frac{1}{2} ln v^{2} + 1 + arctan (v)))

Substitute back

= \frac{1}{6} \cdot 2 (- \frac{1}{4} ln tan^{2} (\frac{6 x}{2}) + 2 tan (\frac{6 x}{2}) - 1 + \frac{1}{2} (\frac{1}{2} ln tan^{2} (\frac{6 x}{2}) + 1 + arctan (tan (\frac{6 x}{2}))))

Simplify

\frac{1}{6} \cdot 2 (- \frac{1}{4} ln tan^{2} (\frac{6 x}{2}) + 2 tan (\frac{6 x}{2}) - 1 + \frac{1}{2} (\frac{1}{2} ln tan^{2} (\frac{6 x}{2}) + 1 + arctan (tan (\frac{6 x}{2})))) : \frac{1}{12} (- ln - 1 + tan^{2} (3 x) + 2 tan (3 x) + ln (sec^{2} (3 x)) + 2 arctan (tan (3 x)))

= \frac{1}{12} (- ln - 1 + tan^{2} (3 x) + 2 tan (3 x) + ln (sec^{2} (3 x)) + 2 arctan (tan (3 x)))

Add a constant to the solution

= \frac{1}{12} (- ln - 1 + tan^{2} (3 x) + 2 tan (3 x) + ln (sec^{2} (3 x)) + 2 arctan (tan (3 x))) + C

x \to 1 lim (x^{3} - 2 x^{2} - 7 x - 6)

x \to \infty lim ((- 1)^{x})

\int \frac{1}{4 t t ^{2} - 16} d t

y^{^{'}} = 2 x^{2}

e^{2 x} + x e^{2 x} + cos (2 x)

What is the integral of 1/(1-tan(6x)) ?
The integral of 1/(1-tan(6x)) is 1/12 (-ln|-1+tan^2(3x)+2tan(3x)|+ln(sec^2(3x))+2arctan(tan(3x)))+C