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

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

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integral of cos^2(2x+1)

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

\frac{1}{2} x - \frac{1}{4} + \frac{1}{8} sin (2 (2 x + 1)) + C

Solution steps

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

Rewrite using trig identities

= \int 1 - sin^{2} (2 x + 1) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d x - \int sin^{2} (2 x + 1) d x

\int 1 d x = x

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

= x - \frac{1}{4} (2 x + 1 - \frac{1}{2} sin (2 (2 x + 1)))

Simplify

x - \frac{1}{4} (2 x + 1 - \frac{1}{2} sin (2 (2 x + 1))) : \frac{1}{2} x - \frac{1}{4} + \frac{1}{8} sin (2 (2 x + 1))

= \frac{1}{2} x - \frac{1}{4} + \frac{1}{8} sin (2 (2 x + 1))

Add a constant to the solution

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

\frac{\partial}{\partial x} (- 2 tan (x) sin (2 y))

\frac{d}{d x} (- A e^{- x})

\frac{d}{d x} (\frac{3 x}{2 x - 1})

\frac{d}{d x} (ln (g (x) (3 x^{2})))

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

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