Question Number 178692 by peter frank last updated on 20/Oct/22
$$\int\frac{\mathrm{tan}\:\left(\mathrm{ln}\:{x}\right).\mathrm{tan}\:\left(\mathrm{ln}\:\frac{{x}}{\mathrm{2}}\right).\mathrm{tan}\:\left(\mathrm{ln}\:\mathrm{2}\right)}{{x}}{dx} \\ $$$$ \\ $$
Answered by mindispower last updated on 21/Oct/22
![tg(ln((x/2))+ln(2))=tg(lnx)=((tg(ln[(2))+tgg(ln((x/2))))/(1−tg(ln(2))tg(ln((x/2))))) ⇔tg(ln(x))tg(ln((x/2)))tgln2=−tgln(2)−tgln(x/2) +tglnx ⇔∫((tgln(x))/x)dx−∫((tgln(2))/x)dx−∫((tg(ln((x/2))))/x)dx =−ln[cos(ln(x)]−tgln(2).ln[x]−2tgln[cos((x/2))]+c](https://www.tinkutara.com/question/Q178767.png)
$${tg}\left({ln}\left(\frac{{x}}{\mathrm{2}}\right)+{ln}\left(\mathrm{2}\right)\right)={tg}\left({lnx}\right)=\frac{{tg}\left({ln}\left[\left(\mathrm{2}\right)\right)+{tgg}\left({ln}\left(\frac{{x}}{\mathrm{2}}\right)\right)\right.}{\mathrm{1}−{tg}\left({ln}\left(\mathrm{2}\right)\right){tg}\left({ln}\left(\frac{{x}}{\mathrm{2}}\right)\right)} \\ $$$$\Leftrightarrow{tg}\left({ln}\left({x}\right)\right){tg}\left({ln}\left(\frac{{x}}{\mathrm{2}}\right)\right){tgln}\mathrm{2}=−{tgln}\left(\mathrm{2}\right)−{tgln}\frac{{x}}{\mathrm{2}} \\ $$$$+{tglnx} \\ $$$$\Leftrightarrow\int\frac{{tgln}\left({x}\right)}{{x}}{dx}−\int\frac{{tgln}\left(\mathrm{2}\right)}{{x}}{dx}−\int\frac{{tg}\left({ln}\left(\frac{{x}}{\mathrm{2}}\right)\right)}{{x}}{dx} \\ $$$$=−{ln}\left[{cos}\left({ln}\left({x}\right)\right]−{tgln}\left(\mathrm{2}\right).{ln}\left[{x}\right]−\mathrm{2}{tgln}\left[{cos}\left(\frac{{x}}{\mathrm{2}}\right)\right]+{c}\right. \\ $$
Commented by mindispower last updated on 21/Oct/22
![i have broken phone sorry [x]=x,x>0,−x other](https://www.tinkutara.com/question/Q178768.png)
$${i}\:{have}\:\:{broken}\:{phone}\:{sorry} \\ $$$$\left[{x}\right]={x},{x}>\mathrm{0},−{x}\:{other} \\ $$
Commented by peter frank last updated on 21/Oct/22
$$\mathrm{thank}\:\mathrm{you}\: \\ $$
Commented by mindispower last updated on 02/Nov/22
$${withe}\:{pleasuf} \\ $$