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

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

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integral of xtan^3(x^2)

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

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

Solution steps

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

Apply u-substitution

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

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

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

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

Add a constant to the solution

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

y^{^{''}} + 6 y^{^{'}} + 6 y = 0

y^{^{'}} = 2 t, y (0) = 1

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

y^{^{'}} = \frac{3 x ^{7} e ^{\frac{y}{x}} + x ^{5} y ^{2}}{x ^{6} y}

x \to 2 lim (\frac{x - 7}{x + 3})

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