Integrate xsin(x) from 0 to 6

\int_{0}^{6} x\cdot \sin\left(x\right)dx

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Answer

$-x\cos\left(x\right)-\frac{19}{68}$

Step by step solution

Problem

$\int_{0}^{6} x\cdot \sin\left(x\right)dx$
1

Use the integration by parts theorem to calculate the integral $\int x\sin\left(x\right)dx$, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
2

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
3

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\sin\left(x\right)dx}\\ \displaystyle{\int dv=\int \sin\left(x\right)dx}\end{matrix}$
4

Solve the integral

$v=\int\sin\left(x\right)dx$
5

Apply the integral of the sine function

$\int_{0}^{6} x\sin\left(x\right)dx$
6

Now replace the values of $u$, $du$ and $v$ in the last formula

$-x\cos\left(x\right)--\int_{0}^{6}\cos\left(x\right)dx$
7

Apply the integral of the cosine function

$-x\cos\left(x\right)-\left[-\sin\left(x\right)\right]_{0}^{6}$
8

Evaluate the definite integral

$\left(\sin\left(6\right)\left(-1\right)-1\cdot \sin\left(0\right)\left(-1\right)\right)\left(-1\right)-x\cos\left(x\right)$
9

Multiply $-1$ times $-1$

$\left(\sin\left(0\right)\cdot 1+\sin\left(6\right)\left(-1\right)\right)\left(-1\right)-x\cos\left(x\right)$
10

Calculating the sine of $6$ degrees

$\left(0\cdot 1-\frac{19}{68}\left(-1\right)\right)\left(-1\right)-x\cos\left(x\right)$
11

Any expression multiplied by $0$ is equal to $0$

$\left(0-\frac{19}{68}\left(-1\right)\right)\left(-1\right)-x\cos\left(x\right)$
12

Multiply $-1$ times $-\frac{19}{68}$

$\left(0+\frac{19}{68}\right)\left(-1\right)-x\cos\left(x\right)$
13

Add the values $\frac{19}{68}$ and $0$

$\frac{19}{68}\left(-1\right)-x\cos\left(x\right)$
14

Multiply $-1$ times $\frac{19}{68}$

$-x\cos\left(x\right)-\frac{19}{68}$

Answer

$-x\cos\left(x\right)-\frac{19}{68}$

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Problem Analysis

Main topic:

Integration by parts

Time to solve it:

0.36 seconds

Views:

96