Question Number 200605 by Calculusboy last updated on 20/Nov/23
Commented by Frix last updated on 21/Nov/23
$$\mathrm{Simply}\:\mathrm{integrate}\:\mathrm{by}\:\mathrm{parts}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{ln}\:\mid\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\mid\:−\frac{{x}}{\mathrm{sin}\:{x}}+{C} \\ $$
Commented by Calculusboy last updated on 21/Nov/23
$$\boldsymbol{{but}}\:\boldsymbol{{sir}},\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{what}}\:\boldsymbol{{i}}\:\boldsymbol{{get}} \\ $$$$\boldsymbol{{In}}\mid\boldsymbol{{cosecx}}−\boldsymbol{{cotx}}\mid−\boldsymbol{{x}}\left(\boldsymbol{{cosxcotx}}+\boldsymbol{{sinx}}\right)+\boldsymbol{{C}} \\ $$
Commented by Frix last updated on 21/Nov/23
![It′s the same. csc x −cot x =((1−cos x)/(sin x))=tan (x/2) [x=2y cos 2y =1−2sin^2 y sin 2y =2cos y sin y ((1−cos 2y)/(sin 2y))=((1−1+2sin^2 y)/(2cos y sin y))=tan y] cos x cot x +sin x =((cos^2 x +sin^2 x)/(sin x))=(1/(sin x))](https://www.tinkutara.com/question/Q200649.png)
$$\mathrm{It}'\mathrm{s}\:\mathrm{the}\:\mathrm{same}. \\ $$$$\mathrm{csc}\:{x}\:−\mathrm{cot}\:{x}\:=\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\left[{x}=\mathrm{2}{y}\right. \\ $$$$\:\:\:\:\:\mathrm{cos}\:\mathrm{2}{y}\:=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{y} \\ $$$$\:\:\:\:\:\mathrm{sin}\:\mathrm{2}{y}\:=\mathrm{2cos}\:{y}\:\mathrm{sin}\:{y} \\ $$$$\left.\:\:\:\:\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{y}}{\mathrm{sin}\:\mathrm{2}{y}}=\frac{\mathrm{1}−\mathrm{1}+\mathrm{2sin}^{\mathrm{2}} \:{y}}{\mathrm{2cos}\:{y}\:\mathrm{sin}\:{y}}=\mathrm{tan}\:{y}\right] \\ $$$$\mathrm{cos}\:{x}\:\mathrm{cot}\:{x}\:+\mathrm{sin}\:{x}\:=\frac{\mathrm{cos}^{\mathrm{2}} \:{x}\:+\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{sin}\:{x}}=\frac{\mathrm{1}}{\mathrm{sin}\:{x}} \\ $$
Commented by Calculusboy last updated on 22/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$