integral of 2cos^2(x) — step-by-step solution

1) Let's compute the integral:

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

We reduce the square using the identity

{\cos (x)}^{2} = \frac{1 + \cos (2 x)}{2}

ⅰ

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

\int (1 + \cos (2 x)) d x

\int (1 + \cos (2 x)) d x

For the math expression:

\cos (2 x)

, since (sin(a))' = a'*cos(a), we try differentiating

\sin (2 x)

2) Let's differentiate:

\frac{d}{d x} (\sin (2 x))

We differentiate using:

\frac{d}{d x} (\sin (f)) = \frac{d}{d x} (f) \cdot \cos (f)

(where f =

2 x

)

ⓘ The derivative of sin(x) is cos(x):

\frac{d}{d x} (\sin (2 x))

\frac{d}{d x} (2 x) \cdot \cos (2 x)

\frac{d}{d x} (2 x) \cdot \cos (2 x)

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (2 x)

2

2 \cdot \cos (2 x)

You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

ⅰ

\int (1 + \cos (2 x)) d x

\frac{1}{2} \cdot \sin (2 x) + \int 1 d x

\frac{1}{2} \cdot \sin (2 x) + \int 1 d x

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{1}{2} \cdot \sin (2 x)

\frac{1 \cdot \sin (2 x)}{2}

\frac{\sin (2 x)}{2}

■The derivative of

x

1

, so

x

is an antiderivative of

1

\int 1 d x

x

\frac{\sin (2 x)}{2} + x

▶Indefinite integrals are defined up to an additive constant C, so this yields:

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