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

The integral of tan^5(3x)sec^3(3x) is (sec^7(3x)}{21}-(2sec^5(3x))/(15)+\frac{sec^3(3x))/9+C

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

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

\frac{sec ^{7} ( 3 x )}{21} - \frac{2 sec ^{5} ( 3 x )}{15} + \frac{sec ^{3} ( 3 x )}{9} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

Expand

\frac{u ^{2} ( u ^{2} - 1 ) ^{2}}{3} : \frac{u ^{6}}{3} - \frac{2 u ^{4}}{3} + \frac{u ^{2}}{3}

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

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

\int \frac{u ^{6}}{3} d u = \frac{u ^{7}}{21}

\int \frac{2 u ^{4}}{3} d u = \frac{2 u ^{5}}{15}

\int \frac{u ^{2}}{3} d u = \frac{u ^{3}}{9}

= \frac{u ^{7}}{21} - \frac{2 u ^{5}}{15} + \frac{u ^{3}}{9}

Substitute back

u = sec (3 x)

= \frac{sec ^{7} ( 3 x )}{21} - \frac{2 sec ^{5} ( 3 x )}{15} + \frac{sec ^{3} ( 3 x )}{9}

Add a constant to the solution

= \frac{sec ^{7} ( 3 x )}{21} - \frac{2 sec ^{5} ( 3 x )}{15} + \frac{sec ^{3} ( 3 x )}{9} + C

What is the integral of tan^5(3x)sec^3(3x) ?
The integral of tan^5(3x)sec^3(3x) is (sec^7(3x)}{21}-(2sec^5(3x))/(15)+\frac{sec^3(3x))/9+C