What is the integral of tan^3(2x+1) ?

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

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integral of tan^3(2x+1)

integral

tan^{3} (2 x + 1)

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

Solution steps

\int tan^{3} (2 x + 1) d x

Apply u-substitution

= \int tan^{3} (u) \frac{1}{2} 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 tan^{3} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

Expand

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

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

\int \frac{1}{v} d v = ln ∣ v ∣

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

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

Substitute back

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

Add a constant to the solution

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

in t e g r a l \frac{- 1}{x + 1 ^{2}}

in t e g r a l \frac{( x + 1 )}{( x - 1 ) ^{2}}

in t e g r a l sin (- u + 3 t)

in t e g r a l - (\frac{( y ^{2} + 1 )}{( 2 x y )})

in t e g r a l \frac{2}{s ^{2}}

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