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

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

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integral of x^2tan^3(x^3)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{3} \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

Apply u-substitution

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

= \frac{1}{3} \cdot \int \frac{- 1 + v ^{2}}{v} d v

Expand

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

= \frac{1}{3} \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}{3} (- \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}{3} (- ln ∣ v ∣ + \frac{v ^{2}}{2})

Substitute back

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

Add a constant to the solution

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

(x^{2} + 4) y^{^{'}} + 3 x y = x, y (0) = 1

9 x^{4} - 9 x^{2}, 8 x^{2}

\int_{0}^{\frac{1}{10}} \frac{1}{( 100 x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\int \frac{( sin ( x ) )}{cos ^{2} ( x )} d x

\frac{d y}{d x}, 9 x y = 5

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