What is the integral of tan^4(x/2)sec^4(x/2) ?

The integral of tan^4(x/2)sec^4(x/2) is 2((tan^5(x/2))/5+(tan^7(x/2))/7)+C

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integral of tan^4(x/2)sec^4(x/2)

\int tan^{4} (\frac{x}{2}) sec^{4} (\frac{x}{2}) d x

2 (\frac{tan ^{5} ( \frac{x}{2} )}{5} + \frac{tan ^{7} ( \frac{x}{2} )}{7}) + C

Solution steps

\int tan^{4} (\frac{x}{2}) sec^{4} (\frac{x}{2}) d x

Apply u-substitution

: \int tan^{4} (u) sec^{4} (u) \cdot 2 d u

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

Rewrite using trig identities

= 2 \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{4} (u) d u

Apply u-substitution

: \int v^{4} (1 + v^{2}) d v

= 2 \cdot \int v^{4} (1 + v^{2}) d v

Expand

v^{4} (1 + v^{2}) : v^{4} + v^{6}

= 2 \cdot \int v^{4} + v^{6} 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^{6} d v)

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

\int v^{6} d v = \frac{v ^{7}}{7}

= 2 (\frac{v ^{5}}{5} + \frac{v ^{7}}{7})

Substitute back

= 2 (\frac{tan ^{5} ( \frac{x}{2} )}{5} + \frac{tan ^{7} ( \frac{x}{2} )}{7})

Add a constant to the solution

= 2 (\frac{tan ^{5} ( \frac{x}{2} )}{5} + \frac{tan ^{7} ( \frac{x}{2} )}{7}) + C

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t \to 0 lim (\frac{4 e ^{t} - 4}{t})

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What is the integral of tan^4(x/2)sec^4(x/2) ?
The integral of tan^4(x/2)sec^4(x/2) is 2((tan^5(x/2))/5+(tan^7(x/2))/7)+C