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

The integral of (cos(2x))/(sin^7(2x)) is -1/12 csc^6(2x)+C

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

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

- \frac{1}{12} csc^{6} (2 x) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{v ^{7}} d v

= \frac{1}{2} \cdot \int \frac{1}{v ^{7}} d v

Apply the Power Rule

: - \frac{1}{6 v ^{6}}

= \frac{1}{2} (- \frac{1}{6 v ^{6}})

Substitute back

= \frac{1}{2} (- \frac{1}{6 sin ^{6} ( 2 x )})

Simplify

\frac{1}{2} (- \frac{1}{6 sin ^{6} ( 2 x )}) : - \frac{1}{12} csc^{6} (2 x)

= - \frac{1}{12} csc^{6} (2 x)

Add a constant to the solution

= - \frac{1}{12} csc^{6} (2 x) + C

\frac{d y}{d x} = x^{3} - 2 y

y = 3 + 4 x^{2} - 2 x^{3}, (1, 5)

f (x) = 3 x^{2} e^{x} + e^{x} (x^{3} + 6)

\frac{d}{d x} (7 sec^{2} (7 x))

\frac{d}{d x} (x^{\frac{5}{x}})

What is the integral of (cos(2x))/(sin^7(2x)) ?
The integral of (cos(2x))/(sin^7(2x)) is -1/12 csc^6(2x)+C