What is the integral of cos^4(5x+2) ?

The integral of cos^4(5x+2) is 1/5 (cos^3(5x+2)sin(5x+2)+3/8 (5x+2-1/4 sin(4(5x+2))))+C

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integral of cos^4(5x+2)

\int cos^{4} (5 x + 2) d x

\frac{1}{5} (cos^{3} (5 x + 2) sin (5 x + 2) + \frac{3}{8} (5 x + 2 - \frac{1}{4} sin (4 (5 x + 2)))) + C

Solution steps

\int cos^{4} (5 x + 2) d x

Apply u-substitution

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

= \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

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

Simplify

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

= \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

= \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))

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

Substitute back

u = 5 x + 2

= \frac{1}{5} (cos^{3} (5 x + 2) sin (5 x + 2) - (- \frac{3}{8} (5 x + 2 - \frac{1}{4} sin (4 (5 x + 2)))))

Simplify

= \frac{1}{5} (cos^{3} (5 x + 2) sin (5 x + 2) + \frac{3}{8} (5 x + 2 - \frac{1}{4} sin (4 (5 x + 2))))

Add a constant to the solution

= \frac{1}{5} (cos^{3} (5 x + 2) sin (5 x + 2) + \frac{3}{8} (5 x + 2 - \frac{1}{4} sin (4 (5 x + 2)))) + C

y^{^{'}} = \frac{( t - y )}{2}

\int cos (x) (6 + 7 sin^{2} (x)) d x

\frac{\partial}{\partial y} (x sin (y))

\frac{d}{d x} (\frac{1}{( 1 + x ^{2} ) ^{2}})

\frac{d y}{d x}, x y = cot (x y)

What is the integral of cos^4(5x+2) ?
The integral of cos^4(5x+2) is 1/5 (cos^3(5x+2)sin(5x+2)+3/8 (5x+2-1/4 sin(4(5x+2))))+C