Calculus Examples

$\int_{0}^{\frac{π}{2}} \cos (x) \sin (\sin (x)) d x$

Step 1

Let $u = \sin (x)$ . Then $d u = \cos (x) d x$ , so $\frac{1}{\cos (x)} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = \sin (x)$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $\sin (x)$ .

$\frac{d}{d x} [\sin (x)]$

Step 1.1.2

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$\cos (x)$

Step 1.2

Substitute the lower limit in for $x$ in $u = \sin (x)$ .

$u_{lower} = \sin (0)$

Step 1.3

The exact value of $\sin (0)$ is $0$ .

$u_{lower} = 0$

Step 1.4

Substitute the upper limit in for $x$ in $u = \sin (x)$ .

$u_{upper} = \sin (\frac{π}{2})$

Step 1.5

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$u_{upper} = 1$

Step 1.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 0$

$u_{upper} = 1$

Step 1.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\int_{0}^{1} \sin (u) d u$

Step 2

The integral of $\sin (u)$ with respect to $u$ is $- \cos (u)$ .

$- \cos (u)]_{0}^{1}$

Step 3

Evaluate $- \cos (u)$ at $1$ and at $0$ .

$- \cos (1) + \cos (0)$

Step 4

The exact value of $\cos (0)$ is $1$ .

$- \cos (1) + 1$

Step 5

Simplify.

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Step 5.1

Evaluate $\cos (1)$ .

$- 1 \cdot 0.5403023 + 1$

Step 5.2

Multiply $- 1$ by $0.5403023$ .

$- 0.5403023 + 1$

Step 5.3

Add $- 0.5403023$ and $1$ .

$0.45969769$