What is the integral of 25cos^2(x) ?

The integral of 25cos^2(x) is 25/2 (x+1/2 sin(2x))+C

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integral of 25cos^2(x)

\int 25 cos^{2} (x) d x

\frac{25}{2} (x + \frac{1}{2} sin (2 x)) + C

Solution steps

\int 25 cos^{2} (x) d x

Take the constant out:

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

= 25 \cdot \int cos^{2} (x) d x

Rewrite using trig identities

= 25 \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

= 25 \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

= 25 \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)

= 25 \cdot \frac{1}{2} (x + \frac{1}{2} sin (2 x))

Simplify

25 \cdot \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : \frac{25}{2} (x + \frac{1}{2} sin (2 x))

= \frac{25}{2} (x + \frac{1}{2} sin (2 x))

Add a constant to the solution

= \frac{25}{2} (x + \frac{1}{2} sin (2 x)) + C

\frac{\partial}{\partial x} (ln (2 x))

y^{^{'}} - \frac{9}{x} y = \frac{y ^{4}}{x ^{12}}

\frac{d y}{d t} = - y^{2} t - y^{2} t^{2}

\frac{d ^{2} y}{d x ^{2}} + \frac{2}{x} \frac{d y}{d x} = 3 x + 4

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

What is the integral of 25cos^2(x) ?
The integral of 25cos^2(x) is 25/2 (x+1/2 sin(2x))+C