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

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

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

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

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

$$I = \sin t$$

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

what did I do wrong?

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

## 3 Answers

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

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

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

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 $I = exp(-\int\dfrac{d u}{u} = exp(-\ln u) = \dfrac{1}{u} = \dfrac{1}{\sin t}$

May be this direct formula helps you

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