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

The integral of tan^3(3x)sec^4(3x) is (sec^6(3x))/(18)-(sec^4(3x))/(12)+C

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

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

\frac{sec ^{6} ( 3 x )}{18} - \frac{sec ^{4} ( 3 x )}{12} + C

Solution steps

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

Rewrite using trig identities

= \int (- 1 + sec^{2} (3 x)) tan (3 x) sec^{4} (3 x) d x

Apply u-substitution

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

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

Expand

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

= \int \frac{u ^{5}}{3} - \frac{u ^{3}}{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 ^{5}}{3} d u - \int \frac{u ^{3}}{3} d u

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

\int \frac{u ^{3}}{3} d u = \frac{u ^{4}}{12}

= \frac{u ^{6}}{18} - \frac{u ^{4}}{12}

Substitute back

u = sec (3 x)

= \frac{sec ^{6} ( 3 x )}{18} - \frac{sec ^{4} ( 3 x )}{12}

Add a constant to the solution

= \frac{sec ^{6} ( 3 x )}{18} - \frac{sec ^{4} ( 3 x )}{12} + C

y^{^{'}} = - 0.02 y - 4

\int - cos (\frac{n x π}{2}) d x

(4 - t) (2 + t^{2})^{- 1}

x y^{^{'}} + y = x^{2} y^{2}

\int \frac{x + 5}{3 x + 9} d x

What is the integral of tan^3(3x)sec^4(3x) ?
The integral of tan^3(3x)sec^4(3x) is (sec^6(3x))/(18)-(sec^4(3x))/(12)+C