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

The integral of (cos^4(x))/(sin^7(x)) is

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

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

\frac{1}{384} (tan^{6} (\frac{x}{2}) - 3 tan^{4} (\frac{x}{2}) - 3 tan^{2} (\frac{x}{2}) + 24 ln tan (\frac{x}{2}) + 3 cot^{2} (\frac{x}{2}) + 3 cot^{4} (\frac{x}{2}) - cot^{6} (\frac{x}{2})) + C

Solution steps

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

Apply u-substitution

: \int \frac{( - u ^{2} + 1 ) ^{4} ( u ^{2} + 1 ) ^{2}}{64 u ^{7}} d u

= \int \frac{( - u ^{2} + 1 ) ^{4} ( u ^{2} + 1 ) ^{2}}{64 u ^{7}} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int \frac{( - u ^{2} + 1 ) ^{4} ( u ^{2} + 1 ) ^{2}}{u ^{7}} d u

Long division

\frac{( - u ^{2} + 1 ) ^{4} ( u ^{2} + 1 ) ^{2}}{u ^{7}} : u^{5} - 2 u^{3} - u + \frac{4 u ^{6} - u ^{4} - 2 u ^{2} + 1}{u ^{7}}

= \frac{1}{64} \cdot \int u^{5} - 2 u^{3} - u + \frac{4 u ^{6} - u ^{4} - 2 u ^{2} + 1}{u ^{7}} 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

= \frac{1}{64} (\int u^{5} d u - \int 2 u^{3} d u - \int u d u + \int \frac{4 u ^{6} - u ^{4} - 2 u ^{2} + 1}{u ^{7}} d u)

\int u^{5} d u = \frac{u ^{6}}{6}

\int 2 u^{3} d u = \frac{u ^{4}}{2}

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

\int \frac{4 u ^{6} - u ^{4} - 2 u ^{2} + 1}{u ^{7}} d u = 4 ln ∣ u ∣ + \frac{1}{2 u ^{2}} + \frac{1}{2 u ^{4}} - \frac{1}{6 u ^{6}}

= \frac{1}{64} (\frac{u ^{6}}{6} - \frac{u ^{4}}{2} - \frac{u ^{2}}{2} + 4 ln ∣ u ∣ + \frac{1}{2 u ^{2}} + \frac{1}{2 u ^{4}} - \frac{1}{6 u ^{6}})

Substitute back

u = tan (\frac{x}{2})

= \frac{1}{64} (\frac{tan ^{6} ( \frac{x}{2} )}{6} - \frac{tan ^{4} ( \frac{x}{2} )}{2} - \frac{tan ^{2} ( \frac{x}{2} )}{2} + 4 ln tan (\frac{x}{2}) + \frac{1}{2 tan ^{2} ( \frac{x}{2} )} + \frac{1}{2 tan ^{4} ( \frac{x}{2} )} - \frac{1}{6 tan ^{6} ( \frac{x}{2} )})

Simplify

\frac{1}{64} (\frac{tan ^{6} ( \frac{x}{2} )}{6} - \frac{tan ^{4} ( \frac{x}{2} )}{2} - \frac{tan ^{2} ( \frac{x}{2} )}{2} + 4 ln tan (\frac{x}{2}) + \frac{1}{2 tan ^{2} ( \frac{x}{2} )} + \frac{1}{2 tan ^{4} ( \frac{x}{2} )} - \frac{1}{6 tan ^{6} ( \frac{x}{2} )}) : \frac{1}{384} (tan^{6} (\frac{x}{2}) - 3 tan^{4} (\frac{x}{2}) - 3 tan^{2} (\frac{x}{2}) + 24 ln tan (\frac{x}{2}) + 3 cot^{2} (\frac{x}{2}) + 3 cot^{4} (\frac{x}{2}) - cot^{6} (\frac{x}{2}))

= \frac{1}{384} (tan^{6} (\frac{x}{2}) - 3 tan^{4} (\frac{x}{2}) - 3 tan^{2} (\frac{x}{2}) + 24 ln tan (\frac{x}{2}) + 3 cot^{2} (\frac{x}{2}) + 3 cot^{4} (\frac{x}{2}) - cot^{6} (\frac{x}{2}))

Add a constant to the solution

= \frac{1}{384} (tan^{6} (\frac{x}{2}) - 3 tan^{4} (\frac{x}{2}) - 3 tan^{2} (\frac{x}{2}) + 24 ln tan (\frac{x}{2}) + 3 cot^{2} (\frac{x}{2}) + 3 cot^{4} (\frac{x}{2}) - cot^{6} (\frac{x}{2})) + C

(1 + x) y^{^{'}} + y = cos (x)

\frac{\partial}{\partial x} (e^{x} sin (3 y))

n = 1 \sum \infty \frac{2}{n !}

\int 5^{3} d x

xg (x)

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