QuestionFind the integral of $\sin x \sin (\cos x)$ with respect to $x$ : $\int \sin x \sin (\cos x) d x$ .

Studdy Solution

STEP 1

Assumptions1. We are asked to integrate the function $\sin x \sin (\cos x)$ with respect to $x$ . . We will use the method of substitution for this integral.

STEP 2

Let's choose a substitution that will simplify the integral. We can let $u = \cos x$ , so that $du = -\sin x \, dx$ .
$u = \cos x$ $du = -\sin x \, dx$

STEP 3

We need to adjust the integral to match our substitution. We can rewrite the integral as $-\int \sin u \, du$

STEP 4

Now, we can integrate $\sin u$ with respect to $u$ .
$-\int \sin u \, du = \cos u + C$

STEP 5

Finally, we substitute back $u = \cos x$ to get the solution in terms of $x$ .
$-\cos (\cos x) + C$ So, the integral of $\sin x \sin (\cos x)$ with respect to $x$ is $-\cos (\cos x) + C$ .

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QuestionFind the integral of $\sin x \sin (\cos x)$ with respect to $x$ : $\int \sin x \sin (\cos x) d x$ .

Studdy Solution

STEP 1

Assumptions1. We are asked to integrate the function $\sin x \sin (\cos x)$ with respect to $x$ . . We will use the method of substitution for this integral.

STEP 2

Let's choose a substitution that will simplify the integral. We can let $u = \cos x$ , so that $du = -\sin x \, dx$ .
$u = \cos x$ $du = -\sin x \, dx$

STEP 3

We need to adjust the integral to match our substitution. We can rewrite the integral as $-\int \sin u \, du$

STEP 4

Now, we can integrate $\sin u$ with respect to $u$ .
$-\int \sin u \, du = \cos u + C$

STEP 5

Finally, we substitute back $u = \cos x$ to get the solution in terms of $x$ .
$-\cos (\cos x) + C$ So, the integral of $\sin x \sin (\cos x)$ with respect to $x$ is $-\cos (\cos x) + C$ .

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QuestionFind the integral of sin⁡xsin⁡(cos⁡x)\sin x \sin (\cos x)sinxsin(cosx) with respect to xxx: ∫sin⁡xsin⁡(cos⁡x)dx\int \sin x \sin (\cos x) d x∫sinxsin(cosx)dx.

Studdy Solution

STEP 1

Assumptions1. We are asked to integrate the function sin⁡xsin⁡(cos⁡x)\sin x \sin (\cos x)sinxsin(cosx) with respect to xxx. . We will use the method of substitution for this integral.

STEP 2

Let's choose a substitution that will simplify the integral. We can let u=cos⁡xu = \cos xu=cosx, so that du=−sin⁡x dxdu = -\sin x \, dxdu=−sinxdx.u=cos⁡xu = \cos xu=cosxdu=−sin⁡x dxdu = -\sin x \, dxdu=−sinxdx

STEP 3

We need to adjust the integral to match our substitution. We can rewrite the integral as−∫sin⁡u du-\int \sin u \, du−∫sinudu

STEP 4

Now, we can integrate sin⁡u\sin usinu with respect to uuu.−∫sin⁡u du=cos⁡u+C-\int \sin u \, du = \cos u + C−∫sinudu=cosu+C

STEP 5

Finally, we substitute back u=cos⁡xu = \cos xu=cosx to get the solution in terms of xxx.−cos⁡(cos⁡x)+C-\cos (\cos x) + C−cos(cosx)+CSo, the integral of sin⁡xsin⁡(cos⁡x)\sin x \sin (\cos x)sinxsin(cosx) with respect to xxx is −cos⁡(cos⁡x)+C-\cos (\cos x) + C−cos(cosx)+C.

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QuestionFind the integral of $\sin x \sin (\cos x)$ with respect to $x$ : $\int \sin x \sin (\cos x) d x$ .

Assumptions1. We are asked to integrate the function $\sin x \sin (\cos x)$ with respect to $x$ . . We will use the method of substitution for this integral.

Let's choose a substitution that will simplify the integral. We can let $u = \cos x$ , so that $du = -\sin x \, dx$ .
$u = \cos x$ $du = -\sin x \, dx$

We need to adjust the integral to match our substitution. We can rewrite the integral as $-\int \sin u \, du$

Now, we can integrate $\sin u$ with respect to $u$ .
$-\int \sin u \, du = \cos u + C$

Finally, we substitute back $u = \cos x$ to get the solution in terms of $x$ .
$-\cos (\cos x) + C$ So, the integral of $\sin x \sin (\cos x)$ with respect to $x$ is $-\cos (\cos x) + C$ .