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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{sec ( u ) tan ( 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 (u) tan (u) d u

Apply u-substitution

: \int 1 d v

= \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot 1 \cdot sec (x^{2})

Simplify

= \frac{1}{2} sec (x^{2})

Add a constant to the solution

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

f (x) = \frac{2 x ^{3} - 4 x ^{2} + 3}{x}

\frac{\partial}{\partial y} (e^{x} tan (y))

\int 4 x^{- 1} - 4^{x} + 10 e^{2 x} - 8 x d x

\frac{d}{d x} ((x - 7)^{2})

\int x^{3} \cdot e^{- x^{2}} d x

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