Menu Close

Sin-Inx-dx-

Tinku Tara 0 Comments
Pin




Question Number 201546 by Calculusboy last updated on 08/Dec/23
∫Sin(Inx)dx
$$\:\:\:\int\boldsymbol{{Sin}}\left(\boldsymbol{{Inx}}\right)\boldsymbol{{dx}} \\ $$
Commented by aleks041103 last updated on 09/Dec/23
is this sine of natural log of x or sth else?
$${is}\:{this}\:{sine}\:{of}\:{natural}\:{log}\:{of}\:{x}\:{or}\:{sth}\:{else}? \\ $$
Commented by Calculusboy last updated on 09/Dec/23
yes it is sine of natural log of x
$$\boldsymbol{{yes}}\:\boldsymbol{{it}}\:\boldsymbol{{is}}\:\boldsymbol{{sine}}\:\boldsymbol{{of}}\:\boldsymbol{{natural}}\:\boldsymbol{{log}}\:\boldsymbol{{of}}\:\boldsymbol{{x}} \\ $$
Commented by mr W last updated on 09/Dec/23
why don′t you make it clear sir to avoid confusing other people again and again? here you mean In(x) is Ln(x), but in Q201229 you said In(x) is not Ln(x), but something else. as i know, natural logarithm is generally written as Ln(x) or Log_e (x) or Log(x), but not as In(x). so please write ln(x) if you mean natural ligarithm, not as In(x).
$${why}\:{don}'{t}\:{you}\:{make}\:{it}\:{clear}\:{sir} \\ $$$${to}\:{avoid}\:{confusing}\:{other}\:{people} \\ $$$${again}\:{and}\:{again}? \\ $$$${here}\:{you}\:{mean}\:\boldsymbol{{In}}\left(\boldsymbol{{x}}\right)\:{is}\:\boldsymbol{{Ln}}\left(\boldsymbol{{x}}\right), \\ $$$${but}\:{in}\:{Q}\mathrm{201229}\:{you}\:{said}\:\boldsymbol{{In}}\left(\boldsymbol{{x}}\right)\:{is} \\ $$$${not}\:\boldsymbol{{Ln}}\left(\boldsymbol{{x}}\right),\:{but}\:{something}\:{else}. \\ $$$${as}\:{i}\:{know},\:{natural}\:{logarithm}\:{is} \\ $$$${generally}\:{written}\:{as}\:{Ln}\left({x}\right)\:{or} \\ $$$${Log}_{{e}} \left({x}\right)\:{or}\:{Log}\left({x}\right),\:{but}\:{not}\:{as}\:{In}\left({x}\right). \\ $$$${so}\:{please}\:{write}\:\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)\:{if}\:{you}\:{mean} \\ $$$${natural}\:{ligarithm},\:{not}\:{as}\:\boldsymbol{{In}}\left(\boldsymbol{{x}}\right). \\ $$
Commented by Calculusboy last updated on 09/Dec/23
okay,am sorry for confusing
$$\boldsymbol{{okay}},\boldsymbol{{am}}\:\boldsymbol{{sorry}}\:\boldsymbol{{for}}\:\boldsymbol{{confusing}} \\ $$
Answered by Mathspace last updated on 09/Dec/23
x=e^t ⇒I=∫ sin(t)e^t dt =Im(∫ e^(it+t) dt) and ∫ e^((1+i)t) dt=(1/(1+i))e^((1+i)t) =((1−i)/2)e^t {cost+isint} =(e^t /2){cost+isint−icost +sint} ⇒I=(e^t /2){sint−cost} +λ =(x/2){sin(lnx)−cos(lnx)}+λ
$${x}={e}^{{t}} \:\Rightarrow{I}=\int\:{sin}\left({t}\right){e}^{{t}} {dt} \\ $$$$={Im}\left(\int\:{e}^{{it}+{t}} {dt}\right) \\ $$$${and}\:\int\:{e}^{\left(\mathrm{1}+{i}\right){t}} {dt}=\frac{\mathrm{1}}{\mathrm{1}+{i}}{e}^{\left(\mathrm{1}+{i}\right){t}} \\ $$$$=\frac{\mathrm{1}−{i}}{\mathrm{2}}{e}^{{t}} \left\{{cost}+{isint}\right\} \\ $$$$=\frac{{e}^{{t}} }{\mathrm{2}}\left\{{cost}+{isint}−{icost}\:+{sint}\right\} \\ $$$$\Rightarrow{I}=\frac{{e}^{{t}} }{\mathrm{2}}\left\{{sint}−{cost}\right\}\:+\lambda \\ $$$$=\frac{{x}}{\mathrm{2}}\left\{{sin}\left({lnx}\right)−{cos}\left({lnx}\right)\right\}+\lambda \\ $$
Answered by JDamian last updated on 09/Dec/23
Calculusboy definitely cannot differentiate a lowercase L from an uppercase I.
$${Calculusboy}\:{definitely}\:\:{cannot}\:{differentiate} \\ $$$${a}\:{lowercase}\:{L}\:{from}\:{an}\:{uppercase}\:{I}. \\ $$
Commented by Calculusboy last updated on 09/Dec/23
i know the difference,i just mistype it when i want to post it
$$\boldsymbol{{i}}\:\boldsymbol{{know}}\:\boldsymbol{{the}}\:\boldsymbol{{difference}},\boldsymbol{{i}}\:\boldsymbol{{just}}\:\boldsymbol{{mistype}}\:\boldsymbol{{it}}\:\boldsymbol{{when}}\:\boldsymbol{{i}}\: \\ $$$$\boldsymbol{{want}}\:\boldsymbol{{to}}\:\boldsymbol{{post}}\:\boldsymbol{{it}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *