What is the integral of 8(tan^3(2sqrt(x)))/(sqrt(x)) ?

The integral of 8(tan^3(2sqrt(x)))/(sqrt(x)) is 8(-ln|sec(2sqrt(x))|+(sec^2(2sqrt(x)))/2)+C

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integral of 8(tan^3(2sqrt(x)))/(sqrt(x))

\int 8 \frac{tan ^{3} ( 2 x )}{x} d x

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

Solution steps

\int 8 \cdot \frac{tan ^{3} ( 2 x )}{x} d x

Take the constant out:

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

= 8 \cdot \int \frac{tan ^{3} ( 2 x )}{x} d x

Apply u-substitution

: \int tan^{3} (u) d u

= 8 \cdot \int tan^{3} (u) d u

Rewrite using trig identities

= 8 \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

Apply u-substitution

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

= 8 \cdot \int \frac{- 1 + v ^{2}}{v} d v

Expand

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

= 8 \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

= 8 (- \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}

= 8 (- ln ∣ v ∣ + \frac{v ^{2}}{2})

Substitute back

= 8 (- ln sec (2 x) + \frac{sec ^{2} ( 2 x )}{2})

Add a constant to the solution

= 8 (- ln sec (2 x) + \frac{sec ^{2} ( 2 x )}{2}) + C

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- x + 5, 0, 2

\int - x e^{x} sin (x) d x

x \to - 2 lim (\frac{x ^{2} - 3}{x + 1})

f (x) = (9 x^{6} + 2 x^{3})^{4}

What is the integral of 8(tan^3(2sqrt(x)))/(sqrt(x)) ?
The integral of 8(tan^3(2sqrt(x)))/(sqrt(x)) is 8(-ln|sec(2sqrt(x))|+(sec^2(2sqrt(x)))/2)+C