What is the integral of 16cos^2(t) ?

The integral of 16cos^2(t) is 8(t+1/2 sin(2t))+C

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integral of 16cos^2(t)

\int 16 cos^{2} (t) d t

8 (t + \frac{1}{2} sin (2 t)) + C

Solution steps

\int 16 cos^{2} (t) d t

Take the constant out:

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

= 16 \cdot \int cos^{2} (t) d t

Rewrite using trig identities

= 16 \cdot \int \frac{1 + cos ( 2 t )}{2} d t

Take the constant out:

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

= 16 \cdot \frac{1}{2} \cdot \int 1 + cos (2 t) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 16 \cdot \frac{1}{2} (\int 1 d t + \int cos (2 t) d t)

\int 1 d t = t

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

= 16 \cdot \frac{1}{2} (t + \frac{1}{2} sin (2 t))

Simplify

16 \cdot \frac{1}{2} (t + \frac{1}{2} sin (2 t)) : 8 (t + \frac{1}{2} sin (2 t))

= 8 (t + \frac{1}{2} sin (2 t))

Add a constant to the solution

= 8 (t + \frac{1}{2} sin (2 t)) + C

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

f (x) = 5 x - 3

\int (x + 5) d x

\frac{d}{d x} (\frac{x ^{3} + 6}{x})

\frac{d}{d x} (\frac{8 + x}{1 - 8 x})

What is the integral of 16cos^2(t) ?
The integral of 16cos^2(t) is 8(t+1/2 sin(2t))+C