What is the integral of xsec^2(1/3)x ?

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

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integral of xsec^2(1/3)x

\int x sec^{2} (\frac{1}{3}) x d x

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

Solution steps

\int x sec^{2} (\frac{1}{3}) x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= sec^{2} (\frac{1}{3}) \cdot \int xx d x

Simplify

xx : x^{2}

= sec^{2} (\frac{1}{3}) \cdot \int x^{2} d x

Apply the Power Rule

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Add a constant to the solution

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

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