What is the integral of sec(1/2)ztan(1/2)z ?

The integral of sec(1/2)ztan(1/2)z is 1/3 sec(1/2)tan(1/2)z^3+C

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integral of sec(1/2)ztan(1/2)z

\int sec (\frac{1}{2}) z tan (\frac{1}{2}) z d z

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

Solution steps

\int sec (\frac{1}{2}) z tan (\frac{1}{2}) z d z

Take the constant out:

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

= sec (\frac{1}{2}) tan (\frac{1}{2}) \cdot \int zz d z

Simplify

zz : z^{2}

= sec (\frac{1}{2}) tan (\frac{1}{2}) \cdot \int z^{2} d z

Apply the Power Rule

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

Simplify

sec (\frac{1}{2}) tan (\frac{1}{2}) \frac{z ^{3}}{3} : \frac{1}{3} sec (\frac{1}{2}) tan (\frac{1}{2}) z^{3}

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

Add a constant to the solution

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

What is the integral of sec(1/2)ztan(1/2)z ?
The integral of sec(1/2)ztan(1/2)z is 1/3 sec(1/2)tan(1/2)z^3+C