What is the integral of 9tsin(t)cos(t) ?

The integral of 9tsin(t)cos(t) is 9/2 (-1/2 tcos(2t)+1/4 sin(2t))+C

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integral of 9tsin(t)cos(t)

\int 9 t sin (t) cos (t) d t

\frac{9}{2} (- \frac{1}{2} t cos (2 t) + \frac{1}{4} sin (2 t)) + C

Solution steps

\int 9 t sin (t) cos (t) d t

Take the constant out:

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

= 9 \cdot \int t sin (t) cos (t) d t

Rewrite using trig identities

= 9 \cdot \int \frac{1}{2} t sin (2 t) d t

Take the constant out:

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

= 9 \cdot \frac{1}{2} \cdot \int t sin (2 t) d t

Apply Integration By Parts

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

= 9 \cdot \frac{1}{2} (- \frac{1}{2} t cos (2 t) - \int - \frac{1}{2} cos (2 t) d t)

\int - \frac{1}{2} cos (2 t) d t = - \frac{1}{4} sin (2 t)

= 9 \cdot \frac{1}{2} (- \frac{1}{2} t cos (2 t) - (- \frac{1}{4} sin (2 t)))

Simplify

9 \cdot \frac{1}{2} (- \frac{1}{2} t cos (2 t) - (- \frac{1}{4} sin (2 t))) : \frac{9}{2} (- \frac{1}{2} t cos (2 t) + \frac{1}{4} sin (2 t))

= \frac{9}{2} (- \frac{1}{2} t cos (2 t) + \frac{1}{4} sin (2 t))

Add a constant to the solution

= \frac{9}{2} (- \frac{1}{2} t cos (2 t) + \frac{1}{4} sin (2 t)) + C

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What is the integral of 9tsin(t)cos(t) ?
The integral of 9tsin(t)cos(t) is 9/2 (-1/2 tcos(2t)+1/4 sin(2t))+C