What is the integral of 4sin(t)cos(t) ?

The integral of 4sin(t)cos(t) is 2sin^2(t)+C

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integral of 4sin(t)cos(t)

\int 4 sin (t) cos (t) d t

2 sin^{2} (t) + C

Solution steps

\int 4 sin (t) cos (t) d t

Take the constant out:

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

= 4 \cdot \int sin (t) cos (t) d t

Apply u-substitution

: \int u d u

= 4 \cdot \int u d u

Apply the Power Rule

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

= 4 \cdot \frac{u ^{2}}{2}

Substitute back

u = sin (t)

= 4 \cdot \frac{sin ^{2} ( t )}{2}

Simplify

4 \cdot \frac{sin ^{2} ( t )}{2} : 2 sin^{2} (t)

= 2 sin^{2} (t)

Add a constant to the solution

= 2 sin^{2} (t) + C

x \to 3 lim (\frac{x ^{3} - 27}{x - 3})

\frac{d y}{d x} = \frac{3 sec ( y )}{( x - 4 ) ^{2}}

\int \frac{2 x ^{5} - x}{x} d x

\frac{d y}{d x} = \frac{( y ^{2} + 6 )}{2 y}

\frac{d}{d x} (cos (x) + sin (y))

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