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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int 2 tan^{3} (u) d u

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= 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

= 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}

= 2 (- ln ∣ v ∣ + \frac{v ^{2}}{2})

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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