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

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

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integral of xsec^3(x^2)

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Reduction

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

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = x^{2}

= \frac{1}{2} (\frac{sec ^{2} ( x ^{2} ) sin ( x ^{2} )}{2} + \frac{1}{2} ln tan (x^{2}) + sec (x^{2}))

Add a constant to the solution

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

\frac{d}{d x} (\frac{4 x - 6}{x - 6})

\frac{d}{d x} (\frac{1}{2} x^{2} e^{x} ln (x))

f (x) = \frac{cos ( e ^{x} )}{x ^{4} + 1}

2 x, \frac{2 x}{3}

\int \frac{x ^{2} + 1}{x 1 + x ^{4}} d x

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