Evaluate the integral by using substitution prior to integration by parts

Integral is:

$\int sin(lnx) dx$

$w = lnx$ .... $dw = \frac 1x$ .... $dx = e^u dw$

Integrating by parts I get

$\int sin(w) dw = sin(w)e^w - \int cos(w)e^w dw$

and I don't know how to go from there. I tried doing integration by parts again but I'm not getting anywhere. Any help is appreciated.

Edit: Integrating by parts again I get:

$e^wsin(w) - e^wcos(w) + \int e^wsin(w) dw$

  • $\begingroup$ Integration by parts again should work. Show your calculation please. $\endgroup$ – mickep Oct 5 '15 at 17:54
  • 1
    $\begingroup$ It works if you do integration by parts twice, calling $ I=\int e^u\sin{u}du$ , and you find $I$ reappears so you can isolate it $\endgroup$ – David Quinn Oct 5 '15 at 17:54
  • $\begingroup$ I'm not understanding. I get how to do the second integration by parts, but I don't know how to integrate $\int e^wsin(w) dw$ ... I don't understand how you get a 1/2 $\endgroup$ – bankey Oct 5 '15 at 18:08

Your integral is

$$\int {\sin (\ln (x))dx} $$

using the substitution $u=ln(x)$ and considering $dx = {e^u}du$ it becomes

$$\int {\sin (u){e^u}du} $$

Now we use integration by parts two times to get

$$\eqalign{ & \int {\sin (u){e^u}du} = \sin (u){e^u} - \int {\cos (u){e^u}du} + C \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sin (u){e^u} - \left( {\cos (u){e^u} - \int { - \sin (u){e^u}du} } \right) + C \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sin (u){e^u} - \cos (u){e^u} - \int {\sin (u){e^u}du} + C \cr} $$

Finally, solve the above equation for $\int {\sin (u){e^u}du} $ which leads to

$$\int {\sin (u){e^u}du} = {1 \over 2}{e^u}\left( {\sin (u) - \cos (u)} \right) + {1 \over 2}C$$

if you want your final answer in $x$ just substitute $u=ln(x)$ to get

$$\int {\sin (\ln (x))dx} = {1 \over 2}x\left( {\sin (\ln (x)) - \cos (\ln (x))} \right) + {1 \over 2}C$$

  • $\begingroup$ How do you integrate $\int sin(u) e^u du$.. I'm not understanding how you get 1/2 $\endgroup$ – bankey Oct 5 '15 at 18:09
  • $\begingroup$ When I integrated twice by parts, the $-\int {\sin (u){e^u}du} $ appeared on the right side. Take it to the left side to get $2\int {\sin (u){e^u}du} $ and then solve for $\int {\sin (u){e^u}du} $. OK? $\endgroup$ – Hosein Rahnama Oct 5 '15 at 18:11
  • $\begingroup$ Ahhhhhh thank you! :) $\endgroup$ – bankey Oct 5 '15 at 18:13

$$\int \sin(\ln x) dx=\int \sin(u)e^u du=\sin(u)e^u-\int \cos(u)e^u du$$ now by doing same method we have, $$\int \cos(u)e^u du=-\sin(u)e^u+\int \sin(u)e^u du$$ combine this two.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.