What is the integral of sec^4((3x)/2)tan^2((3x)/2) ?

The integral of sec^4((3x)/2)tan^2((3x)/2) is 2/3 ((tan^3((3x)/2))/(3)+(tan^5((3x)/2))/(5))+C

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integral of sec^4((3x)/2)tan^2((3x)/2)

\int sec^{4} (\frac{3 x}{2}) tan^{2} (\frac{3 x}{2}) d x

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

Solution steps

\int sec^{4} (\frac{3 x}{2}) tan^{2} (\frac{3 x}{2}) d x

Apply u-substitution

: \int sec^{4} (u) tan^{2} (u) \frac{2}{3} d u

= \int sec^{4} (u) tan^{2} (u) \frac{2}{3} d u

Take the constant out:

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

= \frac{2}{3} \cdot \int sec^{4} (u) tan^{2} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

= \frac{2}{3} \cdot \int v^{2} (1 + v^{2}) d v

Expand

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

= \frac{2}{3} \cdot \int v^{2} + v^{4} 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

= \frac{2}{3} (\int v^{2} d v + \int v^{4} d v)

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

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

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

Substitute back

= \frac{2}{3} (\frac{tan ^{3} ( \frac{3 x}{2} )}{3} + \frac{tan ^{5} ( \frac{3 x}{2} )}{5})

Add a constant to the solution

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

- \frac{e ^{- s}}{( ( s ^{2} + 4 ) \cdot s ^{2} )}

\int 4 x y^{2} + 3 x^{2} d x

y = cos (a^{8} + x^{8})

\frac{d}{d x} (2 csc^{2} (x) cot (x))

\frac{d}{d x} (\frac{4 x ^{\frac{2}{3}}}{x ^{2} + 1})

What is the integral of sec^4((3x)/2)tan^2((3x)/2) ?
The integral of sec^4((3x)/2)tan^2((3x)/2) is 2/3 ((tan^3((3x)/2))/(3)+(tan^5((3x)/2))/(5))+C