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

The integral of tan^4(4t) is 1/12 tan^3(4t)+t-1/4 tan(4t)+C

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integral of tan^4(4t)

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

\frac{1}{12} tan^{3} (4 t) + t - \frac{1}{4} tan (4 t) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

= \frac{1}{4} (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

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

= \frac{1}{4} (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

Substitute back

u = 4 t

= \frac{1}{4} (\frac{tan ^{3} ( 4 t )}{3} - (- 4 t + tan (4 t)))

Simplify

\frac{1}{4} (\frac{tan ^{3} ( 4 t )}{3} - (- 4 t + tan (4 t))) : \frac{1}{12} tan^{3} (4 t) + t - \frac{1}{4} tan (4 t)

= \frac{1}{12} tan^{3} (4 t) + t - \frac{1}{4} tan (4 t)

Add a constant to the solution

= \frac{1}{12} tan^{3} (4 t) + t - \frac{1}{4} tan (4 t) + C

\frac{\partial}{\partial y} (x y + ln (x + y))

y = 2 x e^{x}, (0, 0)

cos (π x)

\frac{d y}{d x} + y cos (x) = 3 cos (x), y (0) = 5

y = 4 x - x^{2}, y = x

What is the integral of tan^4(4t) ?
The integral of tan^4(4t) is 1/12 tan^3(4t)+t-1/4 tan(4t)+C