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

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

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integral of-sin(t)cos(t)

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

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

Solution steps

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

Take the constant out:

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

= - \int sin (t) cos (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

\int \frac{ln ( x ) + 9}{x} d x

\frac{\partial}{\partial x} (2 x^{2} + y)

tan (2 (x + 1))

x \to 1 + lim (\frac{x + 2}{1 - x})

\int_{0}^{10} 100 e^{0.1 x} d x

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