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Problem 10

# Evaluate the integrals. $$\int x \cos \left(-3 x^{2}+4\right) d x$$

Expert verified
The solution to the integral $$\int x \cos \left(-3 x^{2}+4\right) d x$$ is $$-\dfrac{1}{6} \sin(-3x^2+4)+C$$, where $$C$$ is the constant of integration.
See the step by step solution

## Step 1: Identify the substitution

We will substitute $$u = -3x^2 + 4$$. Observe that $$du = -6x dx$$, which can be easily found by taking the derivative of $$u$$ with respect to $$x$$.

## Step 2: Rewrite the integral in terms of the new variable

Now that we have our substitution, let's rewrite the integral in terms of $$u$$: To isolate $$dx$$, we will divide both sides by $$-6x$$: $$dx = \frac{du}{-6x}$$ Now, let's replace the original expression with the one in terms of $$u$$: $$\int x \cos(-3x^2+4) dx = \int x \cos(u) \left(\frac{du}{-6x}\right)$$ Notice that the $$x$$ terms cancel out: $$\int \cos(u) \left(\frac{du}{-6}\right)$$

## Step 3: Integrate with respect to the new variable

Now that we have an expression that is simpler to integrate, let's perform the integration: $$-\dfrac{1}{6}\int \cos(u)du$$ The integral of $$\cos(u)$$ is $$\sin(u)$$, so: $$-\dfrac{1}{6} \sin(u) + C_1$$ (where $$C_1$$ is the constant of integration)

## Step 4: Substitute the original variable back to get the final result

Now, let's substitute the original variable back by replacing $$u$$ with $$-3x^2 + 4$$. So, the final expression becomes: $$-\dfrac{1}{6} \sin(-3x^2+4)+C$$ Where $$C$$ is the constant of integration. Therefore, the solution to the integral is: $$\int x \cos \left(-3 x^{2}+4\right) d x = -\dfrac{1}{6} \sin(-3x^2+4)+C$$

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