Question Number 137083 by mnjuly1970 last updated on 29/Mar/21
$$\:\:\:\:\:\boldsymbol{\phi}=\int^{\:} {sin}\left(\frac{\mathrm{2}}{{x}}\right)\sqrt{\mathrm{1}+{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right)}\:\frac{{dx}}{{x}^{\mathrm{2}} } \\ $$
Answered by Ar Brandon last updated on 29/Mar/21
![β
=β«sin((2/x))(β(1+cos^2 ((1/x))))(dx/x^2 ) u=cos^2 ((1/x)) βdu=(2/x^2 )sin((1/x))cos((1/x))dx β
=β«(β(1+u))du=((2(1+u)^(3/2) )/3)+C =(2/3)[1+cos^2 ((1/x))]^(3/2) +C](Q137085.png)
$$\emptyset=\int\mathrm{sin}\left(\frac{\mathrm{2}}{\mathrm{x}}\right)\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)}\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{u}=\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:\Rightarrow\mathrm{du}=\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{dx} \\ $$$$\emptyset=\int\sqrt{\mathrm{1}+\mathrm{u}}\mathrm{du}=\frac{\mathrm{2}\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{3}/\mathrm{2}} }{\mathrm{3}}+\mathrm{C} \\ $$$$\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\left[\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)\right]^{\mathrm{3}/\mathrm{2}} +\mathrm{C} \\ $$
Commented by mohammad17 last updated on 29/Mar/21
$${put}\:{the}\:{angle}\:{of}\:{sin}\:{is}\:\left(\frac{\mathrm{2}}{{x}}\right) \\ $$$$ \\ $$$${sin}\left(\frac{\mathrm{2}}{{x}}\right)\neq{sin}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 29/Mar/21
$${thanks}\:{alot}.. \\ $$
Commented by Ar Brandon last updated on 29/Mar/21
$$\mathrm{sin}\left(\frac{\mathrm{2}}{\mathrm{x}}\right)=\mathrm{2sin}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$
Commented by Ar Brandon last updated on 29/Mar/21
You're welcome, Sir