What is the integral of ((1+cos(2x))/2)^2 ?

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

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integral of ((1+cos(2x))/2)^2

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

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

Solution steps

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

Simplify

(\frac{1 + cos ( 2 x )}{2})^{2} : cos^{4} (x)

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

Simplify

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

= \int cos^{3} (x) cos (x) d x

Apply Integration By Parts

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

= cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x

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

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

Simplify

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

Add a constant to the solution

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

cos (ln (t))

x \frac{d y}{d x} + 4 y = x^{- 3} e^{x}

\int \frac{x}{2} d x

\frac{\partial}{\partial x} (2 x^{3} y + 3 x y^{2})

\int 2 y^{\frac{3}{2}} d y

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