What is the integral of 7cot^4(5t) ?

The integral of 7cot^4(5t) is -7/15 cot^3(5t)+7t+7/5 cot(5t)+C

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integral of 7cot^4(5t)

\int 7 cot^{4} (5 t) d t

- \frac{7}{15} cot^{3} (5 t) + 7 t + \frac{7}{5} cot (5 t) + C

Solution steps

\int 7 cot^{4} (5 t) d t

Take the constant out:

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

= 7 \cdot \int cot^{4} (5 t) d t

Apply u-substitution

= 7 \cdot \int cot^{4} (u) \frac{1}{5} d u

Take the constant out:

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

= 7 \cdot \frac{1}{5} \cdot \int cot^{4} (u) d u

Apply Integral Reduction

= 7 \cdot \frac{1}{5} (- \frac{cot ^{3} ( u )}{3} - \int cot^{2} (u) d u)

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

= 7 \cdot \frac{1}{5} (- \frac{cot ^{3} ( u )}{3} - (- u - cot (u)))

Substitute back

u = 5 t

= 7 \cdot \frac{1}{5} (- \frac{cot ^{3} ( 5 t )}{3} - (- 5 t - cot (5 t)))

Simplify

7 \cdot \frac{1}{5} (- \frac{cot ^{3} ( 5 t )}{3} - (- 5 t - cot (5 t))) : - \frac{7}{15} cot^{3} (5 t) + 7 t + \frac{7}{5} cot (5 t)

= - \frac{7}{15} cot^{3} (5 t) + 7 t + \frac{7}{5} cot (5 t)

Add a constant to the solution

= - \frac{7}{15} cot^{3} (5 t) + 7 t + \frac{7}{5} cot (5 t) + C

What is the integral of 7cot^4(5t) ?
The integral of 7cot^4(5t) is -7/15 cot^3(5t)+7t+7/5 cot(5t)+C