What is the integral of cos^4(7x+9) ?

The integral of cos^4(7x+9) is 1/7 (cos^3(7x+9)sin(7x+9)+3/8 (7x+9-1/4 sin(4(7x+9))))+C

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integral of cos^4(7x+9)

\int cos^{4} (7 x + 9) d x

\frac{1}{7} (cos^{3} (7 x + 9) sin (7 x + 9) + \frac{3}{8} (7 x + 9 - \frac{1}{4} sin (4 (7 x + 9)))) + C

Solution steps

\int cos^{4} (7 x + 9) d x

Apply u-substitution

: \int cos^{4} (u) \frac{1}{7} d u

= \int cos^{4} (u) \frac{1}{7} d u

Take the constant out:

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

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

Simplify

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

= \frac{1}{7} \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}{7} (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}{7} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = 7 x + 9

= \frac{1}{7} (cos^{3} (7 x + 9) sin (7 x + 9) - (- \frac{3}{8} (7 x + 9 - \frac{1}{4} sin (4 (7 x + 9)))))

Simplify

= \frac{1}{7} (cos^{3} (7 x + 9) sin (7 x + 9) + \frac{3}{8} (7 x + 9 - \frac{1}{4} sin (4 (7 x + 9))))

Add a constant to the solution

= \frac{1}{7} (cos^{3} (7 x + 9) sin (7 x + 9) + \frac{3}{8} (7 x + 9 - \frac{1}{4} sin (4 (7 x + 9)))) + C

\frac{d}{d x} (2 x^{2} e^{2 x})

\frac{x ^{2}}{1 + 8 x}

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

\int_{- \infty}^{0} e^{8 x} d x

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

What is the integral of cos^4(7x+9) ?
The integral of cos^4(7x+9) is 1/7 (cos^3(7x+9)sin(7x+9)+3/8 (7x+9-1/4 sin(4(7x+9))))+C