What is the integral of tan(x/2)sec^3(x/2) ?

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

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integral of tan(x/2)sec^3(x/2)

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

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

Solution steps

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

Apply u-substitution

: \int tan (u) sec^{3} (u) \cdot 2 d u

= \int tan (u) sec^{3} (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int tan (u) sec^{3} (u) d u

Apply u-substitution

: \int \frac{1}{3} d v

= 2 \cdot \int \frac{1}{3} d v

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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