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

The integral of (3tan(x))^3sec(x) is 27(1/3 sec^3(x)-sec(x))+C

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

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

27 (\frac{1}{3} sec^{3} (x) - sec (x)) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{27 sin ^{3} ( x )}{cos ^{4} ( x )} d x

Take the constant out:

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

= 27 \cdot \int \frac{sin ^{3} ( x )}{cos ^{4} ( x )} d x

Rewrite using trig identities

= 27 \cdot \int \frac{( 1 - cos ^{2} ( x ) ) sin ( x )}{cos ^{4} ( x )} d x

Apply u-substitution

: \int - \frac{1 - u ^{2}}{u ^{4}} d u

= 27 \cdot \int - \frac{1 - u ^{2}}{u ^{4}} d u

Expand

- \frac{1 - u ^{2}}{u ^{4}} : - \frac{1}{u ^{4}} + \frac{1}{u ^{2}}

= 27 \cdot \int - \frac{1}{u ^{4}} + \frac{1}{u ^{2}} 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

= 27 (- \int \frac{1}{u ^{4}} d u + \int \frac{1}{u ^{2}} d u)

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

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

= 27 (- (- \frac{1}{3 u ^{3}}) - \frac{1}{u})

Substitute back

u = cos (x)

= 27 (- (- \frac{1}{3 cos ^{3} ( x )}) - \frac{1}{cos ( x )})

Simplify

27 (- (- \frac{1}{3 cos ^{3} ( x )}) - \frac{1}{cos ( x )}) : 27 (\frac{1}{3} sec^{3} (x) - sec (x))

= 27 (\frac{1}{3} sec^{3} (x) - sec (x))

Add a constant to the solution

= 27 (\frac{1}{3} sec^{3} (x) - sec (x)) + C

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What is the integral of (3tan(x))^3sec(x) ?
The integral of (3tan(x))^3sec(x) is 27(1/3 sec^3(x)-sec(x))+C