Find the integral of 1 3 t cos ⁡ ( t ) with respect to t.

Simplify.∫tcos(t)3dtSince&nbsp;13&nbsp;is constant with respect to&nbsp;t, move&nbsp;13&nbsp;out of the integral.13∫tcos(t)dtIntegrate by parts using the formula&nbsp;∫udv=uv-∫vdu, where&nbsp;u=t&nbsp;and&nbsp;dv=cos(t).13(tsin(t)-∫sin(t)dt)The integral of&nbsp;sin(t)&nbsp;with respect to&nbsp;t&nbsp;is&nbsp;-cos(t).13(tsin(t)-(-cos(t)+C))Rewrite&nbsp;13(tsin(t)-(-cos(t)+C))&nbsp;as&nbsp;13(tsin(t)+cos(t))+C.13(tsin(t)+cos(t))+C

Check the solution to "Find the integral of 1 3 t..."

High School
Calculus and Analysis
Calculus 1
Integrals

Iyana Macdonald

Answered question

2022-04-04

Find the integral of

\frac{1}{3} ﻿ t \cos (t)

with respect to t.

Answer & Explanation

Abigailf91er

Beginner2022-04-05Added 13 answers

Simplify.

$\int \frac{t \cos (t)}{3} d t$

Since $\frac{1}{3}$ is constant with respect to $t$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{3} \int t \cos (t) d t$

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = t$ and $d v = \cos (t)$ .

$\frac{1}{3} (t \sin (t) - \int \sin (t) d t)$

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

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

Rewrite $\frac{1}{3} (t \sin (t) - (- \cos (t) + C))$ as $\frac{1}{3} (t \sin (t) + \cos (t)) + C$ .

$\frac{1}{3} (t \sin (t) + \cos (t)) + C$

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