What is the integral of tan^5(x/2)sec^2(x/2) ?

The integral of tan^5(x/2)sec^2(x/2) is 1/3 tan^6(x/2)+C

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integral of tan^5(x/2)sec^2(x/2)

\int tan^{5} (\frac{x}{2}) sec^{2} (\frac{x}{2}) d x

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

Solution steps

\int tan^{5} (\frac{x}{2}) sec^{2} (\frac{x}{2}) d x

Apply u-substitution

: \int 2 u^{5} d u

= \int 2 u^{5} d u

Take the constant out:

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

= 2 \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= 2 \cdot \frac{u ^{6}}{6}

Substitute back

u = tan (\frac{x}{2})

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of tan^5(x/2)sec^2(x/2) ?
The integral of tan^5(x/2)sec^2(x/2) is 1/3 tan^6(x/2)+C