What is the integral of 8cos^2(x) ?

The integral of 8cos^2(x) is 4(x+1/2 sin(2x))+C

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integral of 8cos^2(x)

\int 8 cos^{2} (x) d x

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

Solution steps

\int 8 cos^{2} (x) d x

Take the constant out:

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

= 8 \cdot \int cos^{2} (x) d x

Rewrite using trig identities

= 8 \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

= 8 \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

= 8 \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)

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

Simplify

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

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

Add a constant to the solution

= 4 (x + \frac{1}{2} sin (2 x)) + C

(1 x y^{2} + 4 y) d x + (1 x^{2} y + 4 x) d y = 0

y = x^{2} - 6 x + 9

\frac{d}{d x} (a x^{2} + b x)

\int_{1}^{2} (\frac{1}{8} + \frac{3}{8} x) d x

f (x) = \frac{cos ( x )}{x ^{4}}

What is the integral of 8cos^2(x) ?
The integral of 8cos^2(x) is 4(x+1/2 sin(2x))+C