What is the integral of cos((sqrt(6))/2 t) ?

The integral of cos((sqrt(6))/2 t) is sqrt(2/3)sin((sqrt(3)t)/(sqrt(2)))+C

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integral of cos((sqrt(6))/2 t)

\int cos (\frac{6}{2} t) d t

\frac{2}{3} sin (\frac{3 t}{2}) + C

Solution steps

\int cos (\frac{6}{2} t) d t

Apply u-substitution

: \int cos (u) \frac{2}{3} d u

= \int cos (u) \frac{2}{3} d u

Take the constant out:

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

= \frac{2}{3} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{2}{3} sin (u)

Substitute back

u = \frac{6}{2} t

= \frac{2}{3} sin (\frac{6}{2} t)

Multiply

\frac{6}{2} t : \frac{3 t}{2}

= \frac{2}{3} sin (\frac{3 t}{2})

Add a constant to the solution

= \frac{2}{3} sin (\frac{3 t}{2}) + C

\frac{3}{s ( s ^{2} + 2 s + 5 )}

x \to 1 lim (+ (\frac{4}{2 x - 2}))

5 (x^{3} - x)^{4}

x \to 0 + lim (x ln (tan (2 x)))

\int x (2 sec^{2} (x) + tan^{2} (x)) d x

What is the integral of cos((sqrt(6))/2 t) ?
The integral of cos((sqrt(6))/2 t) is sqrt(2/3)sin((sqrt(3)t)/(sqrt(2)))+C