What is the integral of 9cos^2(x) ?

The integral of 9cos^2(x) is 9/2 (x+1/2 sin(2x))+C

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integral of 9cos^2(x)

\int 9 cos^{2} (x) d x

\frac{9}{2} (x + \frac{1}{2} sin (2 x)) + C

Solution steps

\int 9 cos^{2} (x) d x

Take the constant out:

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

= 9 \cdot \int cos^{2} (x) d x

Rewrite using trig identities

= 9 \cdot \int \frac{1 + cos ( 2 x )}{2} d x

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 1 + cos (2 x) d x

Apply the Sum Rule:

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

= 9 \cdot \frac{1}{2} (\int 1 d x + \int cos (2 x) d x)

\int 1 d x = x

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

= 9 \cdot \frac{1}{2} (x + \frac{1}{2} sin (2 x))

Simplify

9 \cdot \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : \frac{9}{2} (x + \frac{1}{2} sin (2 x))

= \frac{9}{2} (x + \frac{1}{2} sin (2 x))

Add a constant to the solution

= \frac{9}{2} (x + \frac{1}{2} sin (2 x)) + C

\frac{x ^{2} - 3 x}{t ^{2} + 8}

\int \frac{y + 1}{2 y} d y

f (x) = (3 x + 1), at a = 1

f (x) = 4 x - \frac{8}{x}

x \to 1 lim (\frac{7}{1 - x})

What is the integral of 9cos^2(x) ?
The integral of 9cos^2(x) is 9/2 (x+1/2 sin(2x))+C