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Math  /  Calculus

QuestionEvaluate the inderinite integral by using the given substitution to reduce the integral to standard form. ∫sin⁡(2x)dx,u=2x∫sin⁡(2x)dx=□\begin{array}{l} \int \sin (2 x) d x, \quad u=2 x \\ \int \sin (2 x) d x=\square \end{array} □\square

Studdy Solution

STEP 1

1. We are given the integral ∫sin⁡(2x) dx\int \sin(2x) \, dx.
2. We will use the substitution method to simplify the integral.
3. The substitution provided is u=2x u = 2x .

STEP 2

1. Perform the substitution.
2. Change the differential dx dx in terms of du du .
3. Integrate the function in terms of u u .
4. Substitute back in terms of x x .

STEP 3

Perform the substitution u=2x u = 2x . This implies that the differential du du is related to dx dx by differentiating u u with respect to x x :
u=2x u = 2x dudx=2 \frac{du}{dx} = 2 du=2 dx du = 2 \, dx

STEP 4

Solve for dx dx in terms of du du :
dx=du2 dx = \frac{du}{2}
Substitute u=2x u = 2x and dx=du2 dx = \frac{du}{2} into the integral:
∫sin⁡(2x) dx=∫sin⁡(u)⋅du2 \int \sin(2x) \, dx = \int \sin(u) \cdot \frac{du}{2} =12∫sin⁡(u) du = \frac{1}{2} \int \sin(u) \, du

STEP 5

Integrate 12∫sin⁡(u) du\frac{1}{2} \int \sin(u) \, du:
The integral of sin⁡(u)\sin(u) is −cos⁡(u)-\cos(u), so:
12∫sin⁡(u) du=12⋅(−cos⁡(u)) \frac{1}{2} \int \sin(u) \, du = \frac{1}{2} \cdot (-\cos(u)) =−12cos⁡(u)+C = -\frac{1}{2} \cos(u) + C
where C C is the constant of integration.

STEP 6

Substitute back in terms of x x using u=2x u = 2x :
−12cos⁡(u)+C=−12cos⁡(2x)+C -\frac{1}{2} \cos(u) + C = -\frac{1}{2} \cos(2x) + C
The evaluated indefinite integral is:
−12cos⁡(2x)+C \boxed{-\frac{1}{2} \cos(2x) + C}

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