What is the integral of sin^2(2t) ?

The integral of sin^2(2t) is 1/2 (t-1/4 sin(4t))+C

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integral of sin^2(2t)

\int sin^{2} (2 t) d t

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

Solution steps

\int sin^{2} (2 t) d t

Rewrite using trig identities

= \int \frac{1}{2} (1 - cos (4 t)) d t

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 - cos (4 t) d t

Apply the Sum Rule:

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

= \frac{1}{2} (\int 1 d t - \int cos (4 t) d t)

\int 1 d t = t

\int cos (4 t) d t = \frac{1}{4} sin (4 t)

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

Add a constant to the solution

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

\int_{0}^{5} x^{3} 1 + 9 x^{4} d x

\int \frac{4 x + 2}{e ^{x}} d x

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

f (x) = 2 x^{3} + x

\frac{d y}{d x} + y^{7} x + 3 y = 0

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