What is the integral of sec^6(3y)tan^2(3y) ?

The integral of sec^6(3y)tan^2(3y) is 1/3 ((tan^3(3y))/3+(2tan^5(3y))/5+(tan^7(3y))/7)+C

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integral of sec^6(3y)tan^2(3y)

\int sec^{6} (3 y) tan^{2} (3 y) d y

\frac{1}{3} (\frac{tan ^{3} ( 3 y )}{3} + \frac{2 tan ^{5} ( 3 y )}{5} + \frac{tan ^{7} ( 3 y )}{7}) + C

Solution steps

\int sec^{6} (3 y) tan^{2} (3 y) d y

Apply u-substitution

: \int sec^{6} (u) tan^{2} (u) \frac{1}{3} d u

= \int sec^{6} (u) tan^{2} (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 sec^{6} (u) tan^{2} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

v^{2} (1 + v^{2})^{2} : v^{2} + 2 v^{4} + v^{6}

= \frac{1}{3} \cdot \int v^{2} + 2 v^{4} + v^{6} 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 2 v^{4} d v + \int v^{6} d v)

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

\int 2 v^{4} d v = \frac{2 v ^{5}}{5}

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

= \frac{1}{3} (\frac{v ^{3}}{3} + \frac{2 v ^{5}}{5} + \frac{v ^{7}}{7})

Substitute back

= \frac{1}{3} (\frac{tan ^{3} ( 3 y )}{3} + \frac{2 tan ^{5} ( 3 y )}{5} + \frac{tan ^{7} ( 3 y )}{7})

Add a constant to the solution

= \frac{1}{3} (\frac{tan ^{3} ( 3 y )}{3} + \frac{2 tan ^{5} ( 3 y )}{5} + \frac{tan ^{7} ( 3 y )}{7}) + C

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What is the integral of sec^6(3y)tan^2(3y) ?
The integral of sec^6(3y)tan^2(3y) is 1/3 ((tan^3(3y))/3+(2tan^5(3y))/5+(tan^7(3y))/7)+C