What is the integral of t^3(3t^2-4)^{5/2} ?

The integral of t^3(3t^2-4)^{5/2} is 512/9 (1/896 (3t^2-4)^{7/2}+1/4608 (3t^2-4)^{9/2})+C

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integral of t^3(3t^2-4)^{5/2}

\int t^{3} (3 t^{2} - 4)^{\frac{5}{2}} d t

\frac{512}{9} (\frac{1}{896} (3 t^{2} - 4)^{\frac{7}{2}} + \frac{1}{4608} (3 t^{2} - 4)^{\frac{9}{2}}) + C

Solution steps

\int t^{3} (3 t^{2} - 4)^{\frac{5}{2}} d t

Apply Trigonometric Substitution

: \int \frac{512 tan ^{6} ( u ) sec ^{4} ( u )}{9} d u

= \int \frac{512 tan ^{6} ( u ) sec ^{4} ( u )}{9} d u

Take the constant out:

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

= \frac{512}{9} \cdot \int tan^{6} (u) sec^{4} (u) d u

Rewrite using trig identities

= \frac{512}{9} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{6} (u) d u

Apply u-substitution

: \int v^{6} (1 + v^{2}) d v

= \frac{512}{9} \cdot \int v^{6} (1 + v^{2}) d v

Expand

v^{6} (1 + v^{2}) : v^{6} + v^{8}

= \frac{512}{9} \cdot \int v^{6} + v^{8} 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{512}{9} (\int v^{6} d v + \int v^{8} d v)

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

\int v^{8} d v = \frac{v ^{9}}{9}

= \frac{512}{9} (\frac{v ^{7}}{7} + \frac{v ^{9}}{9})

Substitute back

= \frac{512}{9} \frac{tan ^{7} ( arcsec ( \frac{3 ^{\frac{1}{2}}}{2} t ) )}{7} + \frac{tan ^{9} ( arcsec ( \frac{3 ^{\frac{1}{2}}}{2} t ) )}{9}

Simplify

\frac{512}{9} \frac{tan ^{7} ( arcsec ( \frac{3 ^{\frac{1}{2}}}{2} t ) )}{7} + \frac{tan ^{9} ( arcsec ( \frac{3 ^{\frac{1}{2}}}{2} t ) )}{9} : \frac{512}{9} (\frac{1}{896} (3 t^{2} - 4)^{\frac{7}{2}} + \frac{1}{4608} (3 t^{2} - 4)^{\frac{9}{2}})

= \frac{512}{9} (\frac{1}{896} (3 t^{2} - 4)^{\frac{7}{2}} + \frac{1}{4608} (3 t^{2} - 4)^{\frac{9}{2}})

Add a constant to the solution

= \frac{512}{9} (\frac{1}{896} (3 t^{2} - 4)^{\frac{7}{2}} + \frac{1}{4608} (3 t^{2} - 4)^{\frac{9}{2}}) + C

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What is the integral of t^3(3t^2-4)^{5/2} ?
The integral of t^3(3t^2-4)^{5/2} is 512/9 (1/896 (3t^2-4)^{7/2}+1/4608 (3t^2-4)^{9/2})+C