What is the integral from 0 to t of cos^2(2pix) ?

The integral from 0 to t of cos^2(2pix) is 1/2 t+1/(8pi)sin(4pit)

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integral from 0 to t of cos^2(2pix)

\int_{0}^{t} cos^{2} (2 π x) d x

\frac{1}{2} t + \frac{1}{8 π} sin (4 π t)

Solution steps

\int_{0}^{t} cos^{2} (2 π x) d x

Rewrite using trig identities

= \int_{0}^{t} 1 - sin^{2} (2 x π) 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_{0}^{t} 1 d x - \int_{0}^{t} sin^{2} (2 x π) d x

\int_{0}^{t} 1 d x = t

\int_{0}^{t} sin^{2} (2 x π) d x = \frac{1}{2} (t - \frac{1}{4 π} sin (4 π t))

= t - \frac{1}{2} (t - \frac{1}{4 π} sin (4 π t))

Simplify

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

= \frac{1}{2} t + \frac{1}{8 π} sin (4 π t)

\int_{1}^{3} (4^{x + 1} + 2^{x}) d x

\int_{1}^{4} 6 x^{2} - 4 x + 1 d x

\int_{0}^{2} x^{2} ln (x^{2} + 1) d x

\int_{0}^{1} \frac{1}{4} (2 x + y) d y

\int_{1}^{4} x^{2} - 2 x - 1 d x

What is the integral from 0 to t of cos^2(2pix) ?
The integral from 0 to t of cos^2(2pix) is 1/2 t+1/(8pi)sin(4pit)