What is the integral of cot^7(8y) ?

The integral of cot^7(8y) is 1/8 (-1/6 cot^6(8y)+1/4 cot^4(8y)-1/2 cot^2(8y)-ln|sin(8y)|)+C

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integral of cot^7(8y)

\int cot^{7} (8 y) d y

\frac{1}{8} (- \frac{1}{6} cot^{6} (8 y) + \frac{1}{4} cot^{4} (8 y) - \frac{1}{2} cot^{2} (8 y) - ln ∣ sin (8 y) ∣) + C

Solution steps

\int cot^{7} (8 y) d y

Apply u-substitution

: \int cot^{7} (u) \frac{1}{8} d u

= \int cot^{7} (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int cot^{7} (u) d u

Apply Integral Reduction

= \frac{1}{8} (- \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}{8} (- \frac{cot ^{6} ( u )}{6} - (- \frac{1}{4} cot^{4} (u) + \frac{1}{2} cot^{2} (u) + ln ∣ sin (u) ∣))

Substitute back

u = 8 y

= \frac{1}{8} (- \frac{cot ^{6} ( 8 y )}{6} - (- \frac{1}{4} cot^{4} (8 y) + \frac{1}{2} cot^{2} (8 y) + ln ∣ sin (8 y) ∣))

Simplify

\frac{1}{8} (- \frac{cot ^{6} ( 8 y )}{6} - (- \frac{1}{4} cot^{4} (8 y) + \frac{1}{2} cot^{2} (8 y) + ln ∣ sin (8 y) ∣)) : \frac{1}{8} (- \frac{1}{6} cot^{6} (8 y) + \frac{1}{4} cot^{4} (8 y) - \frac{1}{2} cot^{2} (8 y) - ln ∣ sin (8 y) ∣)

= \frac{1}{8} (- \frac{1}{6} cot^{6} (8 y) + \frac{1}{4} cot^{4} (8 y) - \frac{1}{2} cot^{2} (8 y) - ln ∣ sin (8 y) ∣)

Add a constant to the solution

= \frac{1}{8} (- \frac{1}{6} cot^{6} (8 y) + \frac{1}{4} cot^{4} (8 y) - \frac{1}{2} cot^{2} (8 y) - ln ∣ sin (8 y) ∣) + C

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What is the integral of cot^7(8y) ?
The integral of cot^7(8y) is 1/8 (-1/6 cot^6(8y)+1/4 cot^4(8y)-1/2 cot^2(8y)-ln|sin(8y)|)+C