What is the integral of tan^2(3x)sec^4(3x) ?

The integral of tan^2(3x)sec^4(3x) is 1/3 ((tan^3(3x))/3+(tan^5(3x))/5)+C

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integral of tan^2(3x)sec^4(3x)

\int tan^{2} (3 x) sec^{4} (3 x) d x

\frac{1}{3} (\frac{tan ^{3} ( 3 x )}{3} + \frac{tan ^{5} ( 3 x )}{5}) + C

Solution steps

\int tan^{2} (3 x) sec^{4} (3 x) d x

Apply u-substitution

: \int tan^{2} (u) sec^{4} (u) \frac{1}{3} d u

= \int tan^{2} (u) sec^{4} (u) \frac{1}{3} d u

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{3} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{2} (u) d u

Apply u-substitution

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

= \frac{1}{3} \cdot \int v^{2} (1 + v^{2}) d v

Expand

v^{2} (1 + v^{2}) : v^{2} + v^{4}

= \frac{1}{3} \cdot \int v^{2} + v^{4} 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{1}{3} (\int v^{2} d v + \int v^{4} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{3} (\frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= \frac{1}{3} (\frac{tan ^{3} ( 3 x )}{3} + \frac{tan ^{5} ( 3 x )}{5})

Add a constant to the solution

= \frac{1}{3} (\frac{tan ^{3} ( 3 x )}{3} + \frac{tan ^{5} ( 3 x )}{5}) + C

\int (csc^{2} (x) - 1) cot (x) d x

\frac{d}{d x} (\frac{1}{6 x - 7})

x \to - 1 lim (3 x^{2} - 3 x + 3)

\frac{d}{d x} ((- 1)^{n} x^{n})

What is the integral of tan^2(3x)sec^4(3x) ?
The integral of tan^2(3x)sec^4(3x) is 1/3 ((tan^3(3x))/3+(tan^5(3x))/5)+C