What is the integral of 18cot^4(x) ?

The integral of 18cot^4(x) is 18(-1/3 cot^3(x)+x+cot(x))+C

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integral of 18cot^4(x)

\int 18 cot^{4} (x) d x

18 (- \frac{1}{3} cot^{3} (x) + x + cot (x)) + C

Solution steps

\int 18 cot^{4} (x) d x

Take the constant out:

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

= 18 \cdot \int cot^{4} (x) d x

Apply Integral Reduction

= 18 (- \frac{cot ^{3} ( x )}{3} - \int cot^{2} (x) d x)

\int cot^{2} (x) d x = - x - cot (x)

= 18 (- \frac{cot ^{3} ( x )}{3} - (- x - cot (x)))

Simplify

18 (- \frac{cot ^{3} ( x )}{3} - (- x - cot (x))) : 18 (- \frac{1}{3} cot^{3} (x) + x + cot (x))

= 18 (- \frac{1}{3} cot^{3} (x) + x + cot (x))

Add a constant to the solution

= 18 (- \frac{1}{3} cot^{3} (x) + x + cot (x)) + C

\int y^{2} - 1 d y

f (x) = x^{3} - 2 x + 2, (4, 58)

e^{- t} (cos (4 t) - 2 sin (4 t))

\frac{d}{d t} (t^{2} e^{t})

\frac{3}{( x ^{2} - x - 2 )}

What is the integral of 18cot^4(x) ?
The integral of 18cot^4(x) is 18(-1/3 cot^3(x)+x+cot(x))+C