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

The integral of sin(t)2t is 2(-tcos(t)+sin(t))+C

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integral of sin(t)2t

\int sin (t) 2 t d t

2 (- t cos (t) + sin (t)) + C

Solution steps

\int sin (t) \cdot 2 t d t

Take the constant out:

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

= 2 \cdot \int sin (t) t d t

Apply Integration By Parts

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

= 2 (- t cos (t) - \int - cos (t) d t)

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

= 2 (- t cos (t) - (- sin (t)))

Simplify

= 2 (- t cos (t) + sin (t))

Add a constant to the solution

= 2 (- t cos (t) + sin (t)) + C

\int \frac{sin ( 2 x )}{e ^{x}} d x

\frac{d}{d x} (\frac{4 e ^{x}}{2 e ^{x} + 1})

2 sin (2 t)

\frac{d y}{d x} y = \frac{x ^{3}}{3} - x

\frac{\partial}{\partial x} (cos^{6} (x^{8} y^{9}))

What is the integral of sin(t)2t ?
The integral of sin(t)2t is 2(-tcos(t)+sin(t))+C