What is the integral of-cos(t)sin(t) ?

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

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integral of-cos(t)sin(t)

\int - cos (t) sin (t) d t

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

Solution steps

\int - cos (t) sin (t) d t

Take the constant out:

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

= - \int cos (t) sin (t) d t

Apply u-substitution

: \int u d u

= - \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - \frac{u ^{2}}{2}

Substitute back

u = sin (t)

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

Add a constant to the solution

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

(\frac{1}{X})^{^{'}}

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

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

x \to 0 lim (\frac{x ^{4} + 1}{x ^{2}})

x \to 0 lim (∣ sin (x) ∣^{x})

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