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

The integral of xtan^2(x^2)sec^2(x^2) is 1/6 tan^3(x^2)+C

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integral of xtan^2(x^2)sec^2(x^2)

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

\frac{1}{6} tan^{3} (x^{2}) + C

Solution steps

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

Apply u-substitution

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

Rewrite using trig identities

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

Apply u-substitution

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

Apply the Power Rule

= \frac{1}{2} \cdot \frac{v ^{3}}{3}

Substitute back

= \frac{1}{2} \cdot \frac{tan ^{3} ( x ^{2} )}{3}

Simplify

\frac{1}{2} \cdot \frac{tan ^{3} ( x ^{2} )}{3} : \frac{1}{6} tan^{3} (x^{2})

= \frac{1}{6} tan^{3} (x^{2})

Add a constant to the solution

= \frac{1}{6} tan^{3} (x^{2}) + C

What is the integral of xtan^2(x^2)sec^2(x^2) ?
The integral of xtan^2(x^2)sec^2(x^2) is 1/6 tan^3(x^2)+C