Calculus Examples

$\int \sin (t) \sqrt{1 + \cos (t)} d t$

Step 1

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

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Step 1.1

Let $u = 1 + \cos (t)$ . Find $\frac{d u}{d t}$ .

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Step 1.1.1

Differentiate $1 + \cos (t)$ .

$\frac{d}{d t} [1 + \cos (t)]$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $1 + \cos (t)$ with respect to $t$ is $\frac{d}{d t} [1] + \frac{d}{d t} [\cos (t)]$ .

$\frac{d}{d t} [1] + \frac{d}{d t} [\cos (t)]$

Step 1.1.2.2

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [\cos (t)]$

Step 1.1.3

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

$0 - \sin (t)$

Step 1.1.4

Subtract $\sin (t)$ from $0$ .

$- \sin (t)$

Step 1.2

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

$\int - 1 \sqrt{u} d u$

Step 2

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

$- \int \sqrt{u} d u$

Step 3

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{u}$ as $u^{\frac{1}{2}}$ .

$- \int u^{\frac{1}{2}} d u$

Step 4

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

$- (\frac{2}{3} u^{\frac{3}{2}} + C)$

Step 5

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

$- \frac{2}{3} u^{\frac{3}{2}} + C$

Step 6

Replace all occurrences of $u$ with $1 + \cos (t)$ .

$- \frac{2}{3} {(1 + \cos (t))}^{\frac{3}{2}} + C$