242.8.2.8 int x sin (x/5) dx. IBP

[karush]

$\large{242.8.2.8}$
$\displaystyle
I_8=\int(x)\sin{\left(\frac{x}{5}\right)} \, dx=
25\sin\left(\dfrac{x}{5}\right)-5\cos\left(\dfrac{x}{5}\right)x$
$$\begin{align}
u&=\frac{x}{5} &5du&=dx &x&=5u \\
\end{align}\\
$$ thus
$\displaystyle
I_8=25\int u\sin{u} \, du$
IBP
$$\begin{align}
u_1&=u &dv_1&= \sin{u} \, du \\
du_1&=du &v_1&=-\cos{u}
\end{align}$$
Continue?

 

[topsquark]

$\large{242.8.2.8}$
$\displaystyle
I_8=\int(x)\sin{\left(\frac{x}{5}\right)} \, dx=
25\sin\left(\dfrac{x}{5}\right)-5\cos\left(\dfrac{x}{5}\right)x$
$$\begin{align}
u&=\frac{x}{5} &5du&=dx &x&=5u \\
\end{align}\\
$$ thus
$\displaystyle
I_8=25\int u\sin{u} \, du$
IBP
$$\begin{align}
u_1&=u &dv_1&= \sin{u} \, du \\
du_1&=du &v_1&=-\cos{u}
\end{align}$$
Continue?

Integration by parts.
[math]I_8 = -u~cos(u) + \int cos(u)~du[/math]

-Dan

 

[soroban]

$\displaystyle I \;=\; \int(x)\sin{\left(\tfrac{x}{5}\right)} \, dx$

\text{Integration by parts: }\;\int u\,dv \;=\;uv - \int v\,du

. . \begin{array}{cccc}<br /> u \,=\, x &amp;&amp; dv \,=\,\sin(\frac{x}{5})\,dx \\<br /> du \,=\,dx &amp;&amp; v \,=\,-5\cos(\frac{x}{5}) \end{array}

I \;=\;(x)\left(-5\cos(\tfrac{x}{5})\right) - \int\left(-5\cos(\tfrac{x}{5})\right)\,dx

I \;=\;-5x\cos(\tfrac{x}{5}) + 5\int\cos(\tfrac{x}{5})\,dx

I \;=\;-5x\cos(\tfrac{x}{5}) + 25\sin(\tfrac{x}{5}) +C

 

[karush]

$\large{242.8.2.8}$
$\displaystyle
I_8=\int(x)\sin{\left(\frac{x}{5}\right)} \, dx=
25\sin\left(\dfrac{x}{5}\right)-5\cos\left(\dfrac{x}{5}\right)x$
IBP
$\displaystyle \begin{array}{cccc}
u \,=\, x && dv \,=\,\sin(\frac{x}{5})\,dx \\
du \,=\,dx && v \,=\,-5\cos(\frac{x}{5}) \end{array}$
$\displaystyle \;\int u\,dv \;=\;uv - \int v\,du$
$\displaystyle I_8 \;=\;(x)\left(-5\cos(\tfrac{x}{5})\right) - \int\left(-5\cos(\tfrac{x}{5})\right)\,dx$
$\displaystyle I_8 \;=\;-5x\cos(\tfrac{x}{5}) + 5\int\cos(\tfrac{x}{5})\,dx$
$\displaystyle I_8 \;=\;-5x\cos(\tfrac{x}{5}) + 25\sin(\tfrac{x}{5}) +C$

 

[topsquark]

$\large{242.8.2.8}$
$\displaystyle
I_8=\int(x)\sin{\left(\frac{x}{5}\right)} \, dx=
25\sin\left(\dfrac{x}{5}\right)-5\cos\left(\dfrac{x}{5}\right)x$
IBP
$\displaystyle \begin{array}{cccc}
u \,=\, x && dv \,=\,\sin(\frac{x}{5})\,dx \\
du \,=\,dx && v \,=\,-5\cos(\frac{x}{5}) \end{array}$
$\displaystyle \;\int u\,dv \;=\;uv - \int v\,du$
$\displaystyle I_8 \;=\;(x)\left(-5\cos(\tfrac{x}{5})\right) - \int\left(-5\cos(\tfrac{x}{5})\right)\,dx$
$\displaystyle I_8 \;=\;-5x\cos(\tfrac{x}{5}) + 5\int\cos(\tfrac{x}{5})\,dx$
$\displaystyle I_8 \;=\;-5x\cos(\tfrac{x}{5}) + 25\sin(\tfrac{x}{5}) +C$

Looks good. (Nod)

-Dan