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

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

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

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

2 (\frac{sec ^{5} ( \frac{x}{2} )}{5} - \frac{sec ^{3} ( \frac{x}{2} )}{3}) + C

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

= 2 \cdot \int (- 1 + sec^{2} (u)) tan (u) sec^{3} (u) d u

Apply u-substitution

: \int v^{2} (v^{2} - 1) d v

= 2 \cdot \int v^{2} (v^{2} - 1) d v

Expand

v^{2} (v^{2} - 1) : v^{4} - v^{2}

= 2 \cdot \int v^{4} - v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int v^{4} d v - \int v^{2} d v)

\int v^{4} d v = \frac{v ^{5}}{5}

\int v^{2} d v = \frac{v ^{3}}{3}

= 2 (\frac{v ^{5}}{5} - \frac{v ^{3}}{3})

Substitute back

= 2 (\frac{sec ^{5} ( \frac{x}{2} )}{5} - \frac{sec ^{3} ( \frac{x}{2} )}{3})

Add a constant to the solution

= 2 (\frac{sec ^{5} ( \frac{x}{2} )}{5} - \frac{sec ^{3} ( \frac{x}{2} )}{3}) + C

f^{^{'}} (x) = \frac{2 x ^{2} - 9 x + 8}{6 x + 5}

\int r^{3} + 2 r d r

\frac{d}{d t} (2 cos^{2} (t))

\int x (3 x^{2} - 5)^{10} d x

\int x^{4} e^{5 x} d x

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