Math / Calculus

Questionintegral $\int \sin ^{3}(x) e^{\cos (x)} d x$ , we shoulc

Studdy Solution

STEP 1

What is this asking? We need to find the integral of $\sin^3(x) e^{\cos(x)}$ with respect to $x$ . Watch out! Trigonometric integrals can be tricky!
Don't forget your trig identities and be mindful of your signs.

STEP 2

1. Prep the integrand
2. Substitution
3. Expand and integrate
4. Back-substitute

STEP 3

Let's rewrite our integral a bit to make it easier to work with.
We can strategically break up $\sin^3(x)$ into $\sin^2(x) \cdot \sin(x)$ .
This gives us: $\int \sin^2(x) \cdot \sin(x) \cdot e^{\cos(x)} dx$ Why did we do that?
Well, it makes it easier to see a substitution we can make later!

STEP 4

Now, let's use the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$ , which means $\sin^2(x) = 1 - \cos^2(x)$ .
Substituting this into our integral, we get: $\int (1 - \cos^2(x)) \cdot \sin(x) \cdot e^{\cos(x)} dx$ This sets us up perfectly for a u-substitution!

STEP 5

Let $u = \cos(x)$ .
Then, the derivative of $u$ with respect to $x$ is $du/dx = -\sin(x)$ , so $du = -\sin(x) dx$ .
Notice how our integral already has a $\sin(x) dx$ term!
That's why we prepped the integrand the way we did.

STEP 6

Substituting $u$ and $du$ into our integral, we get: $\int (1 - u^2) \cdot e^u \cdot (-du)$ We can pull the negative sign out front: $-\int (1 - u^2) e^u du$

STEP 7

Let's distribute $e^u$ to both terms inside the parentheses: $-\int (e^u - u^2e^u) du$ This allows us to break the integral into two simpler integrals: $-\left[ \int e^u du - \int u^2e^u du \right]$

STEP 8

The integral of $e^u$ with respect to $u$ is simply $e^u$ .
So, we have: $-\left[ e^u - \int u^2e^u du \right]$

STEP 9

The second integral, $\int u^2e^u du$ , requires integration by parts.
Let's do it! Let $v = u^2$ so $dv = 2u du$ , and $dw = e^u du$ so $w = e^u$ .
Then, $\int u^2e^u du = u^2e^u - \int 2ue^u du$ .
We need integration by parts again for $\int 2ue^u du$ .
Let $v = 2u$ so $dv = 2 du$ , and $dw = e^u du$ so $w = e^u$ .
Then, $\int 2ue^u du = 2ue^u - \int 2e^u du = 2ue^u - 2e^u$ .
Putting it all together: $\int u^2e^u du = u^2e^u - (2ue^u - 2e^u) = u^2e^u - 2ue^u + 2e^u$ .

STEP 10

Substituting the results back into our expression: $-\left[ e^u - (u^2e^u - 2ue^u + 2e^u) \right] = -e^u + u^2e^u - 2ue^u + 2e^u = u^2e^u - 2ue^u + e^u$

STEP 11

Finally, let's substitute back $u = \cos(x)$ : $\cos^2(x)e^{\cos(x)} - 2\cos(x)e^{\cos(x)} + e^{\cos(x)}$

STEP 12

The solution to the integral $\int \sin^3(x)e^{\cos(x)} dx$ is $\cos^2(x)e^{\cos(x)} - 2\cos(x)e^{\cos(x)} + e^{\cos(x)} + C$ , where C is the constant of integration.

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