Calculus Examples

$\frac{d y}{d x} = \sin (x) \cos^{2} (x)$

Step 1

Rewrite the equation.

$d y = \sin (x) \cos^{2} (x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int \sin (x) \cos^{2} (x) d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int \sin (x) \cos^{2} (x) d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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Step 2.3.1.1

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

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Step 2.3.1.1.1

Differentiate $\cos (x)$ .

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

Step 2.3.1.1.2

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

$- \sin (x)$

Step 2.3.1.2

Rewrite the problem using $u$ and $d u$ .

$y + C_{1} = \int - 1 u^{2} d u$

Step 2.3.2

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

$y + C_{1} = - \int u^{2} d u$

Step 2.3.3

By the Power Rule, the integral of $u^{2}$ with respect to $u$ is $\frac{1}{3} u^{3}$ .

$y + C_{1} = - (\frac{1}{3} u^{3} + C_{2})$

Step 2.3.4

Rewrite $- (\frac{1}{3} u^{3} + C_{2})$ as $- \frac{1}{3} u^{3} + C_{2}$ .

$y + C_{1} = - \frac{1}{3} u^{3} + C_{2}$

Step 2.3.5

Replace all occurrences of $u$ with $\cos (x)$ .

$y + C_{1} = - \frac{1}{3} \cos^{3} (x) + C_{2}$

Step 2.4

Group the constant of integration on the right side as $K$ .

$y = - \frac{1}{3} \cos^{3} (x) + K$