What is the integral of x^3tan^3(x^4) ?

The integral of x^3tan^3(x^4) is 1/4 (-ln|sec(x^4)|+(sec^2(x^4))/2)+C

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integral of x^3tan^3(x^4)

\int x^{3} tan^{3} (x^{4}) d x

\frac{1}{4} (- ln sec (x^{4}) + \frac{sec ^{2} ( x ^{4} )}{2}) + C

Solution steps

\int x^{3} tan^{3} (x^{4}) d x

Apply u-substitution

: \int \frac{tan ^{3} ( u )}{4} d u

= \int \frac{tan ^{3} ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int tan^{3} (u) d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

Apply u-substitution

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

= \frac{1}{4} \cdot \int \frac{- 1 + v ^{2}}{v} d v

Expand

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

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

Substitute back

= \frac{1}{4} (- ln sec (x^{4}) + \frac{sec ^{2} ( x ^{4} )}{2})

Add a constant to the solution

= \frac{1}{4} (- ln sec (x^{4}) + \frac{sec ^{2} ( x ^{4} )}{2}) + C

f (x) = 3 x

y^{^{'}} + \frac{1}{x + 8} y = x^{- 2}, y (1) = 5

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

\frac{5 + 5 s}{( s ^{2} + 25 ) ^{2}}

f (x) = 4 x - x^{2}, at x = 1

What is the integral of x^3tan^3(x^4) ?
The integral of x^3tan^3(x^4) is 1/4 (-ln|sec(x^4)|+(sec^2(x^4))/2)+C