What is the integral of (cos^6(x))/(sin^{10)(x)} ?

The integral of (cos^6(x))/(sin^{10)(x)} is -1/7 cot^7(x)-1/9 cot^9(x)+C

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integral of (cos^6(x))/(sin^{10)(x)}

\int \frac{cos ^{6} ( x )}{sin ^{10} ( x )} d x

- \frac{1}{7} cot^{7} (x) - \frac{1}{9} cot^{9} (x) + C

Solution steps

\int \frac{cos ^{6} ( x )}{sin ^{10} ( x )} d x

Rewrite using trig identities

= \int \frac{cot ^{6} ( x )}{sin ^{4} ( x )} d x

Use the basic trigonometric identity:

\frac{1}{sin ( x )} = csc (x)

= \int cot^{6} (x) csc^{4} (x) d x

Apply u-substitution

: \int - w^{6} (1 + w^{2}) d w

= \int - w^{6} (1 + w^{2}) d w

Take the constant out:

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

= - \int w^{6} (1 + w^{2}) d w

Expand

w^{6} (1 + w^{2}) : w^{6} + w^{8}

= - \int w^{6} + w^{8} d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int w^{6} d w + \int w^{8} d w)

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

\int w^{8} d w = \frac{w ^{9}}{9}

= - (\frac{w ^{7}}{7} + \frac{w ^{9}}{9})

Substitute back

w = cot (x)

= - (\frac{cot ^{7} ( x )}{7} + \frac{cot ^{9} ( x )}{9})

Simplify

- (\frac{cot ^{7} ( x )}{7} + \frac{cot ^{9} ( x )}{9}) : - \frac{1}{7} cot^{7} (x) - \frac{1}{9} cot^{9} (x)

= - \frac{1}{7} cot^{7} (x) - \frac{1}{9} cot^{9} (x)

Add a constant to the solution

= - \frac{1}{7} cot^{7} (x) - \frac{1}{9} cot^{9} (x) + C

y^{^{''}} + \frac{a _{1^{2}} π ^{2}}{a _{2}^{2}} a_{3}^{2} y = 0

sin (s)

\int - e^{x} + sin (2 x) d x

\int_{0}^{3} cos (\frac{n x π}{3}) d x

\frac{d}{d x} (\frac{x - 1}{x + 2})

What is the integral of (cos^6(x))/(sin^{10)(x)} ?
The integral of (cos^6(x))/(sin^{10)(x)} is -1/7 cot^7(x)-1/9 cot^9(x)+C