What is the integral of cot^3(x)csc^4(x) ?

The integral of cot^3(x)csc^4(x) is -1/6 csc^6(x)+1/4 csc^4(x)+C

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integral of cot^3(x)csc^4(x)

\int cot^{3} (x) csc^{4} (x) d x

- \frac{1}{6} csc^{6} (x) + \frac{1}{4} csc^{4} (x) + C

Solution steps

\int cot^{3} (x) csc^{4} (x) d x

Rewrite using trig identities

= \int \frac{( 1 - sin ^{2} ( x ) ) cos ( x )}{sin ^{7} ( x )} d x

Apply u-substitution

: \int \frac{1 - u ^{2}}{u ^{7}} d u

= \int \frac{1 - u ^{2}}{u ^{7}} d u

Expand

\frac{1 - u ^{2}}{u ^{7}} : \frac{1}{u ^{7}} - \frac{1}{u ^{5}}

= \int \frac{1}{u ^{7}} - \frac{1}{u ^{5}} 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

= \int \frac{1}{u ^{7}} d u - \int \frac{1}{u ^{5}} d u

\int \frac{1}{u ^{7}} d u = - \frac{1}{6 u ^{6}}

\int \frac{1}{u ^{5}} d u = - \frac{1}{4 u ^{4}}

= - \frac{1}{6 u ^{6}} - (- \frac{1}{4 u ^{4}})

Substitute back

u = sin (x)

= - \frac{1}{6 sin ^{6} ( x )} - (- \frac{1}{4 sin ^{4} ( x )})

Simplify

- \frac{1}{6 sin ^{6} ( x )} - (- \frac{1}{4 sin ^{4} ( x )}) : - \frac{1}{6} csc^{6} (x) + \frac{1}{4} csc^{4} (x)

= - \frac{1}{6} csc^{6} (x) + \frac{1}{4} csc^{4} (x)

Add a constant to the solution

= - \frac{1}{6} csc^{6} (x) + \frac{1}{4} csc^{4} (x) + C

f (x) = 4 x - 3

y = 4 e^{- x} + e^{3 x}

\frac{d}{d x} (x^{3} - 6 x^{2} + 9 x - 1)

\frac{d}{d x} (\frac{x - y}{x + y})

\int (x^{2} - 3 x + 5) d x

What is the integral of cot^3(x)csc^4(x) ?
The integral of cot^3(x)csc^4(x) is -1/6 csc^6(x)+1/4 csc^4(x)+C