What is the integral of 4cos^2(t) ?

The integral of 4cos^2(t) is 2t+sin(2t)+C

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integral of 4cos^2(t)

\int 4 cos^{2} (t) d t

2 t + sin (2 t) + C

Solution steps

\int 4 cos^{2} (t) d t

Take the constant out:

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

= 4 \cdot \int cos^{2} (t) d t

Rewrite using trig identities

= 4 \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

= 4 \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

= 4 \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)

= 4 \cdot \frac{1}{2} (t + \frac{1}{2} sin (2 t))

Simplify

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

= 2 t + sin (2 t)

Add a constant to the solution

= 2 t + sin (2 t) + C

x \to 0 lim (\frac{sin ^{2} ( 6 x )}{x})

x \to \infty lim (2 x^{x})

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 9, y^{^{'}} (0) = - 6

n = 1 \sum \infty n (1 - p)^{n - 1}

9 x, x^{2} - 10, 0