What is the integral of 5cot^2(x)csc^4(x) ?

The integral of 5cot^2(x)csc^4(x) is -cot^5(x)-5/3 cot^3(x)+C

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integral of 5cot^2(x)csc^4(x)

\int 5 cot^{2} (x) csc^{4} (x) d x

- cot^{5} (x) - \frac{5}{3} cot^{3} (x) + C

Solution steps

\int 5 cot^{2} (x) csc^{4} (x) d x

Take the constant out:

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

= 5 \cdot \int cot^{2} (x) csc^{4} (x) d x

Apply Trigonometric Substitution

= 5 \cdot \int - u^{2} (u^{2} + 1) d u

Take the constant out:

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

= 5 (- \int u^{2} (u^{2} + 1) d u)

Expand

u^{2} (u^{2} + 1) : u^{4} + u^{2}

= 5 (- \int u^{4} + u^{2} 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

= 5 (- (\int u^{4} d u + \int u^{2} d u))

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

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

= 5 (- (\frac{u ^{5}}{5} + \frac{u ^{3}}{3}))

Substitute back

u = cot (x)

= 5 (- (\frac{cot ^{5} ( x )}{5} + \frac{cot ^{3} ( x )}{3}))

Simplify

5 (- (\frac{cot ^{5} ( x )}{5} + \frac{cot ^{3} ( x )}{3})) : - cot^{5} (x) - \frac{5}{3} cot^{3} (x)

= - cot^{5} (x) - \frac{5}{3} cot^{3} (x)

Add a constant to the solution

= - cot^{5} (x) - \frac{5}{3} cot^{3} (x) + C

What is the integral of 5cot^2(x)csc^4(x) ?
The integral of 5cot^2(x)csc^4(x) is -cot^5(x)-5/3 cot^3(x)+C