What is the integral of (sin^7(2x))/(cos^4(2x)) ?

The integral of (sin^7(2x))/(cos^4(2x)) is 1/6 (sec^3(2x)-9sec(2x)-9cos(2x)+cos^3(2x))+C

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integral of (sin^7(2x))/(cos^4(2x))

\int \frac{sin ^{7} ( 2 x )}{cos ^{4} ( 2 x )} d x

\frac{1}{6} (sec^{3} (2 x) - 9 sec (2 x) - 9 cos (2 x) + cos^{3} (2 x)) + C

Solution steps

\int \frac{sin ^{7} ( 2 x )}{cos ^{4} ( 2 x )} d x

Apply u-substitution

: \int \frac{sin ^{7} ( u )}{2 cos ^{4} ( u )} d u

= \int \frac{sin ^{7} ( u )}{2 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}{2} \cdot \int \frac{sin ^{7} ( u )}{cos ^{4} ( u )} d u

Rewrite using trig identities

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

Substitute back

= \frac{1}{2} (- (- \frac{1}{3 cos ^{3} ( 2 x )}) - \frac{3}{cos ( 2 x )} - 3 cos (2 x) + \frac{cos ^{3} ( 2 x )}{3})

Simplify

\frac{1}{2} (- (- \frac{1}{3 cos ^{3} ( 2 x )}) - \frac{3}{cos ( 2 x )} - 3 cos (2 x) + \frac{cos ^{3} ( 2 x )}{3}) : \frac{1}{6} (sec^{3} (2 x) - 9 sec (2 x) - 9 cos (2 x) + cos^{3} (2 x))

= \frac{1}{6} (sec^{3} (2 x) - 9 sec (2 x) - 9 cos (2 x) + cos^{3} (2 x))

Add a constant to the solution

= \frac{1}{6} (sec^{3} (2 x) - 9 sec (2 x) - 9 cos (2 x) + cos^{3} (2 x)) + C

\frac{\partial}{\partial x} (x y^{3} z^{2})

f (x) = x^{3} - 9 x + 5, at x = 3

y = sec (x^{2} + 3 x)

\int 2 x (x^{2} - 7)^{104} d x

\frac{d}{d x} (15 ln (x) + 4 cos (x) + o)

What is the integral of (sin^7(2x))/(cos^4(2x)) ?
The integral of (sin^7(2x))/(cos^4(2x)) is 1/6 (sec^3(2x)-9sec(2x)-9cos(2x)+cos^3(2x))+C