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

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

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integral of (sin(t))^2

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

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

Solution steps

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

Simplify

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

Rewrite using trig identities

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

\int 1 d t = t

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (e^{2 x} + e^{- 2 x})

x \to 0 lim (\frac{3 ^{x} - 1}{x})

f (x) = \frac{- 4 x ^{2} + 3 x - 7}{2 x ^{2} - 10 x - 3}

\int \frac{1}{x ( x ^{2} + 5 )} d x

\int - e^{- k t} k M d t

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