What is the integral of (sin^7(6x))/(cos^4(6x)) ?

The integral of (sin^7(6x))/(cos^4(6x)) is 1/18 (sec^3(6x)-9sec(6x)-9cos(6x)+cos^3(6x))+C

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integral of (sin^7(6x))/(cos^4(6x))

\int \frac{sin ^{7} ( 6 x )}{cos ^{4} ( 6 x )} d x

\frac{1}{18} (sec^{3} (6 x) - 9 sec (6 x) - 9 cos (6 x) + cos^{3} (6 x)) + C

Solution steps

\int \frac{sin ^{7} ( 6 x )}{cos ^{4} ( 6 x )} d x

Apply u-substitution

: \int \frac{sin ^{7} ( u )}{6 cos ^{4} ( u )} d u

= \int \frac{sin ^{7} ( u )}{6 cos ^{4} ( u )} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{sin ^{7} ( u )}{cos ^{4} ( u )} d u

Rewrite using trig identities

= \frac{1}{6} \cdot \int \frac{( 1 - cos ^{2} ( u ) ) ^{3} sin ( u )}{cos ^{4} ( u )} d u

Apply u-substitution

: \int - \frac{( 1 - v ^{2} ) ^{3}}{v ^{4}} d v

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

Expand

- \frac{( 1 - v ^{2} ) ^{3}}{v ^{4}} : - \frac{1}{v ^{4}} + \frac{3}{v ^{2}} - 3 + v^{2}

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

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

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

\int 3 d v = 3 v

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

= \frac{1}{6} (- (- \frac{1}{3 v ^{3}}) - \frac{3}{v} - 3 v + \frac{v ^{3}}{3})

Substitute back

= \frac{1}{6} (- (- \frac{1}{3 cos ^{3} ( 6 x )}) - \frac{3}{cos ( 6 x )} - 3 cos (6 x) + \frac{cos ^{3} ( 6 x )}{3})

Simplify

\frac{1}{6} (- (- \frac{1}{3 cos ^{3} ( 6 x )}) - \frac{3}{cos ( 6 x )} - 3 cos (6 x) + \frac{cos ^{3} ( 6 x )}{3}) : \frac{1}{18} (sec^{3} (6 x) - 9 sec (6 x) - 9 cos (6 x) + cos^{3} (6 x))

= \frac{1}{18} (sec^{3} (6 x) - 9 sec (6 x) - 9 cos (6 x) + cos^{3} (6 x))

Add a constant to the solution

= \frac{1}{18} (sec^{3} (6 x) - 9 sec (6 x) - 9 cos (6 x) + cos^{3} (6 x)) + C

x \to 2.9 lim (x^{2} + 3 x)

y = \frac{2 t}{3 + t ^{2}}

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

\frac{\partial}{\partial y} (e^{x y} x)

\frac{( - 10 x - 9 y x ^{2} )}{3 ( x ^{3} - 5 y ^{2} )}

What is the integral of (sin^7(6x))/(cos^4(6x)) ?
The integral of (sin^7(6x))/(cos^4(6x)) is 1/18 (sec^3(6x)-9sec(6x)-9cos(6x)+cos^3(6x))+C