What is the integral of sec^3(x/4)tan(x/4) ?

The integral of sec^3(x/4)tan(x/4) is 4/3 sec^3(x/4)+C

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integral of sec^3(x/4)tan(x/4)

\int sec^{3} (\frac{x}{4}) tan (\frac{x}{4}) d x

\frac{4}{3} sec^{3} (\frac{x}{4}) + C

Solution steps

\int sec^{3} (\frac{x}{4}) tan (\frac{x}{4}) d x

Apply u-substitution

= \int sec^{3} (u) tan (u) \cdot 4 d u

Take the constant out:

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

= 4 \cdot \int sec^{3} (u) tan (u) d u

Apply u-substitution

= 4 \cdot \int \frac{1}{3} d v

Integral of a constant:

\int a d x = a x

= 4 \cdot \frac{1}{3} v

Substitute back

= 4 \cdot \frac{1}{3} sec^{3} (\frac{x}{4})

Simplify

4 \cdot \frac{1}{3} sec^{3} (\frac{x}{4}) : \frac{4}{3} sec^{3} (\frac{x}{4})

= \frac{4}{3} sec^{3} (\frac{x}{4})

Add a constant to the solution

= \frac{4}{3} sec^{3} (\frac{x}{4}) + C

What is the integral of sec^3(x/4)tan(x/4) ?
The integral of sec^3(x/4)tan(x/4) is 4/3 sec^3(x/4)+C