What is the integral of cot^7(2x) ?

The integral of cot^7(2x) is 1/2 (-1/6 cot^6(2x)+1/4 cot^4(2x)-1/2 cot^2(2x)-ln|sin(2x)|)+C

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integral of cot^7(2x)

\int cot^{7} (2 x) d x

\frac{1}{2} (- \frac{1}{6} cot^{6} (2 x) + \frac{1}{4} cot^{4} (2 x) - \frac{1}{2} cot^{2} (2 x) - ln ∣ sin (2 x) ∣) + C

Solution steps

\int cot^{7} (2 x) d x

Apply u-substitution

: \int cot^{7} (u) \frac{1}{2} d u

= \int cot^{7} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int cot^{7} (u) d u

Apply Integral Reduction

= \frac{1}{2} (- \frac{cot ^{6} ( u )}{6} - \int cot^{5} (u) d u)

\int cot^{5} (u) d u = - \frac{1}{4} cot^{4} (u) + \frac{1}{2} cot^{2} (u) + ln ∣ sin (u) ∣

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

Substitute back

u = 2 x

= \frac{1}{2} (- \frac{cot ^{6} ( 2 x )}{6} - (- \frac{1}{4} cot^{4} (2 x) + \frac{1}{2} cot^{2} (2 x) + ln ∣ sin (2 x) ∣))

Simplify

\frac{1}{2} (- \frac{cot ^{6} ( 2 x )}{6} - (- \frac{1}{4} cot^{4} (2 x) + \frac{1}{2} cot^{2} (2 x) + ln ∣ sin (2 x) ∣)) : \frac{1}{2} (- \frac{1}{6} cot^{6} (2 x) + \frac{1}{4} cot^{4} (2 x) - \frac{1}{2} cot^{2} (2 x) - ln ∣ sin (2 x) ∣)

= \frac{1}{2} (- \frac{1}{6} cot^{6} (2 x) + \frac{1}{4} cot^{4} (2 x) - \frac{1}{2} cot^{2} (2 x) - ln ∣ sin (2 x) ∣)

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{6} cot^{6} (2 x) + \frac{1}{4} cot^{4} (2 x) - \frac{1}{2} cot^{2} (2 x) - ln ∣ sin (2 x) ∣) + C

\frac{d y}{d x} = \frac{x - 1}{3 y ^{2}}

3 e^{5 x} cos (6 x)

\int \frac{x ^{4} e ^{x} - 6 x ^{3}}{x ^{4}} d x

\frac{2 s}{s + 10}

y = x^{3} + 2, (2, 2)

What is the integral of cot^7(2x) ?
The integral of cot^7(2x) is 1/2 (-1/6 cot^6(2x)+1/4 cot^4(2x)-1/2 cot^2(2x)-ln|sin(2x)|)+C