Math / Calculus

QuestionPrevious Problem Problem List Next Problem (1 point) Evaluate the integral. $\int 6 x \sin (-3 x) d x$
Answer = $\square$ $+C$

Studdy Solution

STEP 1

What is this asking? We need to find the indefinite integral of $6x \cdot \sin(-3x)$ with respect to $x$ . Watch out! Don't forget the constant of integration, $+C$ , and be careful with those negative signs!

STEP 2

1. Simplify the integrand
2. Integrate by parts
3. Simplify the result

STEP 3

Hey everyone!
Let's simplify this integral first.
Remember that $\sin(-u) = -\sin(u)$ .
So, we can rewrite our integral as: $\int 6x \cdot \sin(-3x) \, dx = \int 6x \cdot (-\sin(3x)) \, dx = -\int 6x \cdot \sin(3x) \, dx$ Much cleaner, right?

STEP 4

Now, let's integrate by parts!
Remember the formula: $\int u \, dv = uv - \int v \, du$ .

STEP 5

Let's choose $u = 6x$ , so $du = 6 \, dx$ .
Then, $dv = \sin(3x) \, dx$ .
To find $v$ , we integrate $dv$ : $v = \int \sin(3x) \, dx = -\frac{1}{3}\cos(3x)$ We divide by 3 because the chain rule would bring a 3 out front if we were differentiating.

STEP 6

Now, plug everything into the integration by parts formula: \begin{align} -\int 6x \cdot \sin(3x) \, dx &= -(6x \cdot \left( -\frac{1}{3}\cos(3x) \right) - \int -\frac{1}{3}\cos(3x) \cdot 6 \, dx) \\ &= -(-2x\cos(3x) + 2\int \cos(3x) \, dx)\end{align}

STEP 7

Let's finish the last integral: $2\int \cos(3x) \, dx = 2 \cdot \frac{1}{3}\sin(3x) = \frac{2}{3}\sin(3x)$ Again, we divide by 3 to account for the chain rule.

STEP 8

Putting it all together, we get: $-(-2x\cos(3x) + \frac{2}{3}\sin(3x)) = 2x\cos(3x) - \frac{2}{3}\sin(3x)$

STEP 9

Our final answer is: $2x\cos(3x) - \frac{2}{3}\sin(3x) + C$

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QuestionPrevious Problem Problem List Next Problem (1 point) Evaluate the integral. $\int 6 x \sin (-3 x) d x$
Answer = $\square$ $+C$

Studdy Solution

STEP 1

What is this asking? We need to find the indefinite integral of $6x \cdot \sin(-3x)$ with respect to $x$ . Watch out! Don't forget the constant of integration, $+C$ , and be careful with those negative signs!

STEP 2

1. Simplify the integrand
2. Integrate by parts
3. Simplify the result

STEP 3

Hey everyone!
Let's simplify this integral first.
Remember that $\sin(-u) = -\sin(u)$ .
So, we can rewrite our integral as: $\int 6x \cdot \sin(-3x) \, dx = \int 6x \cdot (-\sin(3x)) \, dx = -\int 6x \cdot \sin(3x) \, dx$ Much cleaner, right?

STEP 4

Now, let's integrate by parts!
Remember the formula: $\int u \, dv = uv - \int v \, du$ .

STEP 5

Let's choose $u = 6x$ , so $du = 6 \, dx$ .
Then, $dv = \sin(3x) \, dx$ .
To find $v$ , we integrate $dv$ : $v = \int \sin(3x) \, dx = -\frac{1}{3}\cos(3x)$ We divide by 3 because the chain rule would bring a 3 out front if we were differentiating.

STEP 6

Now, plug everything into the integration by parts formula: \begin{align} -\int 6x \cdot \sin(3x) \, dx &= -(6x \cdot \left( -\frac{1}{3}\cos(3x) \right) - \int -\frac{1}{3}\cos(3x) \cdot 6 \, dx) \\ &= -(-2x\cos(3x) + 2\int \cos(3x) \, dx)\end{align}

STEP 7

Let's finish the last integral: $2\int \cos(3x) \, dx = 2 \cdot \frac{1}{3}\sin(3x) = \frac{2}{3}\sin(3x)$ Again, we divide by 3 to account for the chain rule.

STEP 8

Putting it all together, we get: $-(-2x\cos(3x) + \frac{2}{3}\sin(3x)) = 2x\cos(3x) - \frac{2}{3}\sin(3x)$

STEP 9

Our final answer is: $2x\cos(3x) - \frac{2}{3}\sin(3x) + C$

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QuestionPrevious Problem Problem List Next Problem (1 point) Evaluate the integral. ∫6xsin⁡(−3x)dx\int 6 x \sin (-3 x) d x∫6xsin(−3x)dxAnswer = □\square□ +C+C+C

Studdy Solution

STEP 1

What is this asking? We need to find the *indefinite integral* of 6x⋅sin⁡(−3x)6x \cdot \sin(-3x)6x⋅sin(−3x) with respect to xxx. Watch out! Don't forget the constant of integration, +C+C+C, and be careful with those negative signs!

STEP 2

1. Simplify the integrand2. Integrate by parts3. Simplify the result

STEP 3

STEP 4

Now, let's **integrate by parts**!Remember the formula: ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du∫udv=uv−∫vdu.

STEP 5

STEP 6

Now, plug everything into the integration by parts formula: \begin{align*} -\int 6x \cdot \sin(3x) \, dx &= -(6x \cdot \left( -\frac{1}{3}\cos(3x) \right) - \int -\frac{1}{3}\cos(3x) \cdot 6 \, dx) \\ &= -(-2x\cos(3x) + 2\int \cos(3x) \, dx)\end{align*}

STEP 7

Let's finish the last integral: 2∫cos⁡(3x) dx=2⋅13sin⁡(3x)=23sin⁡(3x)2\int \cos(3x) \, dx = 2 \cdot \frac{1}{3}\sin(3x) = \frac{2}{3}\sin(3x)2∫cos(3x)dx=2⋅31​sin(3x)=32​sin(3x) Again, we divide by **3** to account for the chain rule.

STEP 8

Putting it all together, we get: −(−2xcos⁡(3x)+23sin⁡(3x))=2xcos⁡(3x)−23sin⁡(3x) -(-2x\cos(3x) + \frac{2}{3}\sin(3x)) = 2x\cos(3x) - \frac{2}{3}\sin(3x) −(−2xcos(3x)+32​sin(3x))=2xcos(3x)−32​sin(3x)

STEP 9

Our final answer is: 2xcos⁡(3x)−23sin⁡(3x)+C 2x\cos(3x) - \frac{2}{3}\sin(3x) + C 2xcos(3x)−32​sin(3x)+C

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QuestionPrevious Problem Problem List Next Problem (1 point) Evaluate the integral. $\int 6 x \sin (-3 x) d x$
Answer = $\square$ $+C$

What is this asking? We need to find the indefinite integral of $6x \cdot \sin(-3x)$ with respect to $x$ . Watch out! Don't forget the constant of integration, $+C$ , and be careful with those negative signs!

1. Simplify the integrand
2. Integrate by parts
3. Simplify the result

Now, let's integrate by parts!
Remember the formula: $\int u \, dv = uv - \int v \, du$ .

Now, plug everything into the integration by parts formula: \begin{align} -\int 6x \cdot \sin(3x) \, dx &= -(6x \cdot \left( -\frac{1}{3}\cos(3x) \right) - \int -\frac{1}{3}\cos(3x) \cdot 6 \, dx) \\ &= -(-2x\cos(3x) + 2\int \cos(3x) \, dx)\end{align}

Let's finish the last integral: $2\int \cos(3x) \, dx = 2 \cdot \frac{1}{3}\sin(3x) = \frac{2}{3}\sin(3x)$ Again, we divide by 3 to account for the chain rule.

Putting it all together, we get: $-(-2x\cos(3x) + \frac{2}{3}\sin(3x)) = 2x\cos(3x) - \frac{2}{3}\sin(3x)$

Our final answer is: $2x\cos(3x) - \frac{2}{3}\sin(3x) + C$