What is the integral of-36sin(6t-pi/2) ?

The integral of-36sin(6t-pi/2) is 6sin(6t)+C

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integral of-36sin(6t-pi/2)

\int - 36 sin (6 t - \frac{π}{2}) d t

6 sin (6 t) + C

Solution steps

\int - 36 sin (6 t - \frac{π}{2}) d t

Take the constant out:

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

= - 36 \cdot \int sin (6 t - \frac{π}{2}) d t

Rewrite using trig identities

= - 36 \cdot \int - cos (6 t) d t

Take the constant out:

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

= - 36 (- \int cos (6 t) d t)

Apply u-substitution

: \int cos (u) \frac{1}{6} d u

= - 36 (- \int cos (u) \frac{1}{6} d u)

Take the constant out:

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

= - 36 (- \frac{1}{6} \cdot \int cos (u) d u)

Use the common integral:

\int cos (u) d u = sin (u)

= - 36 (- \frac{1}{6} sin (u))

Substitute back

u = 6 t

= - 36 (- \frac{1}{6} sin (6 t))

Simplify

- 36 (- \frac{1}{6} sin (6 t)) : 6 sin (6 t)

= 6 sin (6 t)

Add a constant to the solution

= 6 sin (6 t) + C

\int_{0}^{1} \frac{1}{1 + x} d x

2 9 + x^{2}

\int_{0}^{2 π} sin (x) d x

\int 3 sin (3 x) + 2 cos (4 x) d x

y^{^{'''}} + 2 y^{^{''}} - y^{^{'}} - 2 y = sin (3 t)

What is the integral of-36sin(6t-pi/2) ?
The integral of-36sin(6t-pi/2) is 6sin(6t)+C