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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - u^{2} d u

= \int - u^{2} d u

Take the constant out:

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

= - \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= - \frac{u ^{3}}{3}

Substitute back

u = cot (x)

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

Add a constant to the solution

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

\int 7 x^{3} e^{x^{4}} d x

\frac{d}{d x} ((4 x^{2} - 5)^{3})

x y^{^{'}} - 2 y - 8 = 0

f (x) = 3 x (18 x^{4} + \frac{13}{x + 1})

\frac{d}{d x} (\frac{6 x ^{2} + 6 x + 6}{x})

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