What is the integral of 4cos^4(5x) ?

The integral of 4cos^4(5x) is 4/5 (cos^3(5x)sin(5x)+3/8 (5x-1/4 sin(20x)))+C

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integral of 4cos^4(5x)

\int 4 cos^{4} (5 x) d x

\frac{4}{5} (cos^{3} (5 x) sin (5 x) + \frac{3}{8} (5 x - \frac{1}{4} sin (20 x))) + C

Solution steps

\int 4 cos^{4} (5 x) d x

Take the constant out:

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

= 4 \cdot \int cos^{4} (5 x) d x

Apply u-substitution

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

= 4 \cdot \int cos^{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

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

Simplify

cos^{4} (u) : cos^{3} (u) cos (u)

= 4 \cdot \frac{1}{5} \cdot \int cos^{3} (u) cos (u) d u

Apply Integration By Parts

: cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u

= 4 \cdot \frac{1}{5} (cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u)

\int - 3 cos^{2} (u) sin^{2} (u) d u = - \frac{3}{8} (u - \frac{1}{4} sin (4 u))

= 4 \cdot \frac{1}{5} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = 5 x

= 4 \cdot \frac{1}{5} (cos^{3} (5 x) sin (5 x) - (- \frac{3}{8} (5 x - \frac{1}{4} sin (4 \cdot 5 x))))

Simplify

4 \cdot \frac{1}{5} (cos^{3} (5 x) sin (5 x) - (- \frac{3}{8} (5 x - \frac{1}{4} sin (4 \cdot 5 x)))) : \frac{4}{5} (cos^{3} (5 x) sin (5 x) + \frac{3}{8} (5 x - \frac{1}{4} sin (20 x)))

= \frac{4}{5} (cos^{3} (5 x) sin (5 x) + \frac{3}{8} (5 x - \frac{1}{4} sin (20 x)))

Add a constant to the solution

= \frac{4}{5} (cos^{3} (5 x) sin (5 x) + \frac{3}{8} (5 x - \frac{1}{4} sin (20 x))) + C

f (x) = \frac{e ^{x}}{1 + e ^{x}}

f (x) = \frac{x ^{2} + 2 x - 2}{( x + 1 ) ^{2}}

\frac{d}{d t} (6 cos (t))

\int_{1}^{2} x^{- 2} d x

y = \frac{16}{x ^{2} - 6 x - 55}, x = - 3, x = 3

What is the integral of 4cos^4(5x) ?
The integral of 4cos^4(5x) is 4/5 (cos^3(5x)sin(5x)+3/8 (5x-1/4 sin(20x)))+C