What is the integral of tan^3(3x)sec^6(3x) ?

The integral of tan^3(3x)sec^6(3x) is (sec^8(3x))/(24)-(sec^6(3x))/(18)+C

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integral of tan^3(3x)sec^6(3x)

\int tan^{3} (3 x) sec^{6} (3 x) d x

\frac{sec ^{8} ( 3 x )}{24} - \frac{sec ^{6} ( 3 x )}{18} + C

Solution steps

\int tan^{3} (3 x) sec^{6} (3 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (3 x)) tan (3 x) sec^{6} (3 x) d x

Apply u-substitution

: \int \frac{w ^{5} ( w ^{2} - 1 )}{3} d w

= \int \frac{w ^{5} ( w ^{2} - 1 )}{3} d w

Expand

\frac{w ^{5} ( w ^{2} - 1 )}{3} : \frac{w ^{7}}{3} - \frac{w ^{5}}{3}

= \int \frac{w ^{7}}{3} - \frac{w ^{5}}{3} d w

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{w ^{7}}{3} d w - \int \frac{w ^{5}}{3} d w

\int \frac{w ^{7}}{3} d w = \frac{w ^{8}}{24}

\int \frac{w ^{5}}{3} d w = \frac{w ^{6}}{18}

= \frac{w ^{8}}{24} - \frac{w ^{6}}{18}

Substitute back

w = sec (3 x)

= \frac{sec ^{8} ( 3 x )}{24} - \frac{sec ^{6} ( 3 x )}{18}

Add a constant to the solution

= \frac{sec ^{8} ( 3 x )}{24} - \frac{sec ^{6} ( 3 x )}{18} + C

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What is the integral of tan^3(3x)sec^6(3x) ?
The integral of tan^3(3x)sec^6(3x) is (sec^8(3x))/(24)-(sec^6(3x))/(18)+C