What is the integral of 3tan^3(2x)sec^5(2x) ?

The integral of 3tan^3(2x)sec^5(2x) is 3((sec^7(2x))/(14)-(sec^5(2x))/(10))+C

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integral of 3tan^3(2x)sec^5(2x)

\int 3 tan^{3} (2 x) sec^{5} (2 x) d x

3 (\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10}) + C

Solution steps

\int 3 tan^{3} (2 x) sec^{5} (2 x) d x

Take the constant out:

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

= 3 \cdot \int tan^{3} (2 x) sec^{5} (2 x) d x

Rewrite using trig identities

= 3 \cdot \int (- 1 + sec^{2} (2 x)) tan (2 x) sec^{5} (2 x) d x

Apply u-substitution

: \int \frac{u ^{4} ( u ^{2} - 1 )}{2} d u

= 3 \cdot \int \frac{u ^{4} ( u ^{2} - 1 )}{2} d u

Expand

\frac{u ^{4} ( u ^{2} - 1 )}{2} : \frac{u ^{6}}{2} - \frac{u ^{4}}{2}

= 3 \cdot \int \frac{u ^{6}}{2} - \frac{u ^{4}}{2} d u

Apply the Sum Rule:

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

= 3 (\int \frac{u ^{6}}{2} d u - \int \frac{u ^{4}}{2} d u)

\int \frac{u ^{6}}{2} d u = \frac{u ^{7}}{14}

\int \frac{u ^{4}}{2} d u = \frac{u ^{5}}{10}

= 3 (\frac{u ^{7}}{14} - \frac{u ^{5}}{10})

Substitute back

u = sec (2 x)

= 3 (\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10})

Add a constant to the solution

= 3 (\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10}) + C

x^{3} \cdot y^{^{'}} = x

[(z - 1)^{3} (z + 1)^{2}]

\frac{\partial}{\partial y} (y + (cos (x))^{2})

((\frac{1 + e ^{x}}{2})^{9})^{^{'}}

What is the integral of 3tan^3(2x)sec^5(2x) ?
The integral of 3tan^3(2x)sec^5(2x) is 3((sec^7(2x))/(14)-(sec^5(2x))/(10))+C