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

The integral of xsec^2(x^2)tan(x^2) is 1/4 tan^2(x^2)+C

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

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

\frac{1}{4} tan^{2} (x^{2}) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int v d v

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

Apply the Power Rule

: \frac{v ^{2}}{2}

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{tan ^{2} ( x ^{2} )}{2} : \frac{1}{4} tan^{2} (x^{2})

= \frac{1}{4} tan^{2} (x^{2})

Add a constant to the solution

= \frac{1}{4} tan^{2} (x^{2}) + C

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What is the integral of xsec^2(x^2)tan(x^2) ?
The integral of xsec^2(x^2)tan(x^2) is 1/4 tan^2(x^2)+C