0
$\begingroup$

$$I = \exp(-\int\frac{\cos t}{\sin t}dt)$$

$$I = \exp(-\int\frac{1}{\sin t}\cos tdt)$$

$$I = \exp(-\int\frac{1}{\sin^2 t}dt)$$

$$I = \exp(-\ln|\sin^2t|)$$

$$I = \sin t$$

but answer should be $$\color{red}{\frac{1}{\sin t}}$$

what did I do wrong?

$\endgroup$
  • $\begingroup$ yes It should be, I'll fix that $\endgroup$ – Reboot Feb 3 '17 at 6:54
1
$\begingroup$

How does $\frac 1{\sin t} \cos t$ become $\frac 1{\sin^2 t}$??

$-\int \frac {\cos t}{\sin t} dt\\ u = \sin t\\ du = \cos t\\ -\int \frac 1u du\\ -\ln u\\ -\ln \sin t\\ \ln (\sin t)^{-1}\\ \ln\frac {1}{\sin t}\\ e^{\ln\frac {1}{\sin t}} = \frac {1}{\sin t}$

$\endgroup$
  • $\begingroup$ $cost=\frac{1}{sint}$ $\endgroup$ – Reboot Feb 3 '17 at 6:55
  • $\begingroup$ @Reboot Where did you hear that? That is very false. $\endgroup$ – user223391 Feb 3 '17 at 6:56
  • $\begingroup$ $\csc t = \frac 1{\sin t}$ $\endgroup$ – Doug M Feb 3 '17 at 6:56
  • $\begingroup$ My apologies, I used the wrong identity. Thanks for awesome explanation $\endgroup$ – Reboot Feb 3 '17 at 6:57
1
$\begingroup$

The mistake lies on the third step $I = exp(-\int \dfrac{\cos t}{\sin t} d t $ From $u$-substition let $u = \sin t$ then $d u = \cos t d t$ hence we have $I = exp(-\int\dfrac{d u}{u} = exp(-\ln u) = \dfrac{1}{u} = \dfrac{1}{\sin t}$

$\endgroup$
1
$\begingroup$

May be this direct formula helps you

$$\int\frac{f'(x)}{f(x)}dx=\ln\lvert{f(x)}\rvert+c$$

$\endgroup$

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.