What is the integral of tan^6(9x)sec^4(9x) ?

The integral of tan^6(9x)sec^4(9x) is 1/9 ((tan^7(9x))/7+(tan^9(9x))/9)+C

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integral of tan^6(9x)sec^4(9x)

\int tan^{6} (9 x) sec^{4} (9 x) d x

\frac{1}{9} (\frac{tan ^{7} ( 9 x )}{7} + \frac{tan ^{9} ( 9 x )}{9}) + C

Solution steps

\int tan^{6} (9 x) sec^{4} (9 x) d x

Apply u-substitution

: \int tan^{6} (u) sec^{4} (u) \frac{1}{9} d u

= \int tan^{6} (u) sec^{4} (u) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int tan^{6} (u) sec^{4} (u) d u

Rewrite using trig identities

= \frac{1}{9} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{6} (u) d u

Apply u-substitution

: \int v^{6} (1 + v^{2}) d v

= \frac{1}{9} \cdot \int v^{6} (1 + v^{2}) d v

Expand

v^{6} (1 + v^{2}) : v^{6} + v^{8}

= \frac{1}{9} \cdot \int v^{6} + v^{8} 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}{9} (\int v^{6} d v + \int v^{8} d v)

\int v^{6} d v = \frac{v ^{7}}{7}

\int v^{8} d v = \frac{v ^{9}}{9}

= \frac{1}{9} (\frac{v ^{7}}{7} + \frac{v ^{9}}{9})

Substitute back

= \frac{1}{9} (\frac{tan ^{7} ( 9 x )}{7} + \frac{tan ^{9} ( 9 x )}{9})

Add a constant to the solution

= \frac{1}{9} (\frac{tan ^{7} ( 9 x )}{7} + \frac{tan ^{9} ( 9 x )}{9}) + C

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What is the integral of tan^6(9x)sec^4(9x) ?
The integral of tan^6(9x)sec^4(9x) is 1/9 ((tan^7(9x))/7+(tan^9(9x))/9)+C