Calculus Examples

\int t \sin (t) d t

$\int t \sin (t) d t$

Step 1

Integrate by parts using the formula

\int u d v = u v - \int v d u

$\int u d v = u v - \int v d u$ , where

u = t

$u = t$ and

d v = \sin (t)

$d v = \sin (t)$ .

t (- \cos (t)) - \int - \cos (t) d t

$t (- \cos (t)) - \int - \cos (t) d t$

Step 2

Since

- 1

$- 1$ is constant with respect to

t

$t$ , move

- 1

$- 1$ out of the integral.

t (- \cos (t)) - - \int \cos (t) d t

$t (- \cos (t)) - - \int \cos (t) d t$

Step 3

Simplify.

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Step 3.1

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

t (- \cos (t)) + 1 \int \cos (t) d t

$t (- \cos (t)) + 1 \int \cos (t) d t$

Step 3.2

Multiply

\int \cos (t) d t

$\int \cos (t) d t$ by

1

$1$ .

t (- \cos (t)) + \int \cos (t) d t

$t (- \cos (t)) + \int \cos (t) d t$

t (- \cos (t)) + \int \cos (t) d t

$t (- \cos (t)) + \int \cos (t) d t$

Step 4

The integral of

\cos (t)

$\cos (t)$ with respect to

t

$t$ is

\sin (t)

$\sin (t)$ .

t (- \cos (t)) + \sin (t) + C

$t (- \cos (t)) + \sin (t) + C$

Step 5

Rewrite

t (- \cos (t)) + \sin (t) + C

$t (- \cos (t)) + \sin (t) + C$ as

- t \cos (t) + \sin (t) + C

$- t \cos (t) + \sin (t) + C$ .

- t \cos (t) + \sin (t) + C

$- t \cos (t) + \sin (t) + C$

\int_{}^{} t \sin (t) d t

$\int_{}^{} t \sin (t) d t$

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[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$