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

The integral of tan^4(5t) is 1/15 tan^3(5t)+t-1/5 tan(5t)+C

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integral of tan^4(5t)

\int tan^{4} (5 t) d t

\frac{1}{15} tan^{3} (5 t) + t - \frac{1}{5} tan (5 t) + C

Solution steps

\int tan^{4} (5 t) d t

Apply u-substitution

: \int tan^{4} (u) \frac{1}{5} d u

= \int tan^{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

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

Apply Integral Reduction

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

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

= \frac{1}{5} (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

Substitute back

u = 5 t

= \frac{1}{5} (\frac{tan ^{3} ( 5 t )}{3} - (- 5 t + tan (5 t)))

Simplify

\frac{1}{5} (\frac{tan ^{3} ( 5 t )}{3} - (- 5 t + tan (5 t))) : \frac{1}{15} tan^{3} (5 t) + t - \frac{1}{5} tan (5 t)

= \frac{1}{15} tan^{3} (5 t) + t - \frac{1}{5} tan (5 t)

Add a constant to the solution

= \frac{1}{15} tan^{3} (5 t) + t - \frac{1}{5} tan (5 t) + C

\int \frac{1 + 3 x}{( 1 - x ) ( 3 x - 5 )} d x

\int e^{7 t} d t

y = 5 + 5 x^{2} - 2 x^{3}, (2, 9)

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

(0, - 2), (- 2, - 2)

What is the integral of tan^4(5t) ?
The integral of tan^4(5t) is 1/15 tan^3(5t)+t-1/5 tan(5t)+C