What is the integral of 4cos(2θ-5) ?

The integral of 4cos(2θ-5) is 2sin(2θ-5)+C

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integral of 4cos(2θ-5)

\int 4 cos (2 θ - 5) d θ

2 sin (2 θ - 5) + C

Solution steps

\int 4 cos (2 θ - 5) d θ

Take the constant out:

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

= 4 \cdot \int cos (2 θ - 5) d θ

Rewrite using trig identities

= 4 \cdot \int cos (2 θ) cos (5) + sin (2 θ) sin (5) d θ

Apply the Sum Rule:

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

= 4 (\int cos (2 θ) cos (5) d θ + \int sin (2 θ) sin (5) d θ)

\int cos (2 θ) cos (5) d θ = cos (5) \frac{1}{2} sin (2 θ)

\int sin (2 θ) sin (5) d θ = - \frac{1}{2} sin (5) cos (2 θ)

= 4 (cos (5) \frac{1}{2} sin (2 θ) - \frac{1}{2} sin (5) cos (2 θ))

Simplify

4 (cos (5) \frac{1}{2} sin (2 θ) - \frac{1}{2} sin (5) cos (2 θ)) : 2 sin (2 θ - 5)

= 2 sin (2 θ - 5)

Add a constant to the solution

= 2 sin (2 θ - 5) + C

\int x + ln (x) d x

x \to 0 + lim (- x + 1)

y^{^{'}} = \frac{y}{x} + (\frac{y}{x})^{2}

x \to - 4 lim (5 x - 6)