Math  /  Calculus

QuestionВычислить неопределенные интегралы, используя метод интегрирования по частям:
1. 2xcosxdx\int 2 x \cos x d x

Studdy Solution

STEP 1

1. The problem requires the use of integration by parts.
2. The formula for integration by parts is udv=uvvdu\int u \, dv = uv - \int v \, du.
3. We need to choose u u and dv dv appropriately from the integrand.

STEP 2

1. Identify u u and dv dv .
2. Differentiate u u to find du du .
3. Integrate dv dv to find v v .
4. Apply the integration by parts formula.
5. Simplify and solve the resulting integral.

STEP 3

Identify u u and dv dv from the integral 2xcosxdx\int 2x \cos x \, dx.
Let: u=2x u = 2x dv=cosxdx dv = \cos x \, dx

STEP 4

Differentiate u u to find du du :
du=ddx(2x)dx=2dx du = \frac{d}{dx}(2x) \, dx = 2 \, dx

STEP 5

Integrate dv dv to find v v :
v=cosxdx=sinx v = \int \cos x \, dx = \sin x

STEP 6

Apply the integration by parts formula udv=uvvdu\int u \, dv = uv - \int v \, du:
2xcosxdx=2xsinxsinx2dx \int 2x \cos x \, dx = 2x \sin x - \int \sin x \cdot 2 \, dx

STEP 7

Simplify the expression:
=2xsinx2sinxdx = 2x \sin x - 2 \int \sin x \, dx

STEP 8

Solve the remaining integral sinxdx\int \sin x \, dx:
sinxdx=cosx \int \sin x \, dx = -\cos x
Substitute back into the expression:
2xsinx2(cosx) 2x \sin x - 2(-\cos x)
=2xsinx+2cosx+C = 2x \sin x + 2\cos x + C
where C C is the constant of integration.
The indefinite integral is:
2xsinx+2cosx+C \boxed{2x \sin x + 2\cos x + C}

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