Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 97361 by student work last updated on 07/Jun/20

∫((xdx)/(sin^2 x−3))=?

$$\int\frac{\mathrm{xdx}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}−\mathrm{3}}=? \\ $$

Commented by student work last updated on 07/Jun/20

who is intellagent?

$$\mathrm{who}\:\mathrm{is}\:\mathrm{intellagent}? \\ $$

Commented by mr W last updated on 07/Jun/20

i don′t know who is intelligent.  but according to this question of you,  i know at least who is not intelligent.

$${i}\:{don}'{t}\:{know}\:{who}\:{is}\:{intelligent}. \\ $$$${but}\:{according}\:{to}\:{this}\:{question}\:{of}\:{you}, \\ $$$${i}\:{know}\:{at}\:{least}\:{who}\:{is}\:{not}\:{intelligent}. \\ $$

Commented by som(math1967) last updated on 07/Jun/20

����

Commented by student work last updated on 07/Jun/20

solve

$$\mathrm{solve} \\ $$

Commented by student work last updated on 07/Jun/20

who can?

$$\mathrm{who}\:\mathrm{can}? \\ $$

Commented by Tinku Tara last updated on 07/Jun/20

Hi student work  please stop putting comments  such as who is intelligent, solve  who can etc.  Polite comments like pls help,  i need help for assignment  are fine.

$$\mathrm{Hi}\:\mathrm{student}\:\mathrm{work} \\ $$$$\mathrm{please}\:\mathrm{stop}\:\mathrm{putting}\:\mathrm{comments} \\ $$$$\mathrm{such}\:\mathrm{as}\:\mathrm{who}\:\mathrm{is}\:\mathrm{intelligent},\:\mathrm{solve} \\ $$$$\mathrm{who}\:\mathrm{can}\:\mathrm{etc}. \\ $$$$\mathrm{Polite}\:\mathrm{comments}\:\mathrm{like}\:\mathrm{pls}\:\mathrm{help}, \\ $$$$\mathrm{i}\:\mathrm{need}\:\mathrm{help}\:\mathrm{for}\:\mathrm{assignment} \\ $$$$\mathrm{are}\:\mathrm{fine}. \\ $$

Answered by mathmax by abdo last updated on 07/Jun/20

I =∫  ((xdx)/(sin^2 x−3)) =∫  ((xdx)/(−2−cos^2 x)) =−∫  ((xdx)/(2+(1/(1+tan^2 x))))  =−∫  ((x(1+tan^2 x))/(3+2tan^2 x))dx  changement tanx =u give  I =−∫  (((1+u^2 )arctanu)/(3+2u^2 )) (du/(1+u^2 )) =−∫  ((arctanu)/(3+2u^2 )) du  =−(1/3) ∫  ((arctanu)/(1+(2/3)u^2 )) du   and changement  (√(2/3))u =z give  I =−(1/3) ∫  ((arctan((√(3/2))z))/(1+z^2 ))×(√(3/2))dz =−(1/(√6)) ∫  ((arctan((√(3/2))z))/(1+z^2 ))dz  let put  f(α) =∫_0 ^x  ((arctan(αz))/(1+z^2 )) dz⇒  f^′ (α) =∫_0 ^x  (z/((1+z^2 )(1+α^2 z^2 ))) dz  =_(αz =u)    ∫_0 ^(αx)  (u/(α(1+(u^2 /α^2 ))(1+u^2 ))) (du/α)  =∫_0 ^(αx)  ((udu)/((u^2  +α^2 )(u^2  +1)))  this integral is solvable by decomposition  ...be continued....

$$\mathrm{I}\:=\int\:\:\frac{\mathrm{xdx}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}−\mathrm{3}}\:=\int\:\:\frac{\mathrm{xdx}}{−\mathrm{2}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:=−\int\:\:\frac{\mathrm{xdx}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}} \\ $$$$=−\int\:\:\frac{\mathrm{x}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{3}+\mathrm{2tan}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{tanx}\:=\mathrm{u}\:\mathrm{give} \\ $$$$\mathrm{I}\:=−\int\:\:\frac{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\mathrm{arctanu}}{\mathrm{3}+\mathrm{2u}^{\mathrm{2}} }\:\frac{\mathrm{du}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\:=−\int\:\:\frac{\mathrm{arctanu}}{\mathrm{3}+\mathrm{2u}^{\mathrm{2}} }\:\mathrm{du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\:\int\:\:\frac{\mathrm{arctanu}}{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\mathrm{u}^{\mathrm{2}} }\:\mathrm{du}\:\:\:\mathrm{and}\:\mathrm{changement}\:\:\sqrt{\frac{\mathrm{2}}{\mathrm{3}}}\mathrm{u}\:=\mathrm{z}\:\mathrm{give} \\ $$$$\mathrm{I}\:=−\frac{\mathrm{1}}{\mathrm{3}}\:\int\:\:\frac{\mathrm{arctan}\left(\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\mathrm{z}\right)}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }×\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\mathrm{dz}\:=−\frac{\mathrm{1}}{\sqrt{\mathrm{6}}}\:\int\:\:\frac{\mathrm{arctan}\left(\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\mathrm{z}\right)}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }\mathrm{dz} \\ $$$$\mathrm{let}\:\mathrm{put}\:\:\mathrm{f}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\mathrm{x}} \:\frac{\mathrm{arctan}\left(\alpha\mathrm{z}\right)}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }\:\mathrm{dz}\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\alpha\right)\:=\int_{\mathrm{0}} ^{\mathrm{x}} \:\frac{\mathrm{z}}{\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)\left(\mathrm{1}+\alpha^{\mathrm{2}} \mathrm{z}^{\mathrm{2}} \right)}\:\mathrm{dz}\:\:=_{\alpha\mathrm{z}\:=\mathrm{u}} \:\:\:\int_{\mathrm{0}} ^{\alpha\mathrm{x}} \:\frac{\mathrm{u}}{\alpha\left(\mathrm{1}+\frac{\mathrm{u}^{\mathrm{2}} }{\alpha^{\mathrm{2}} }\right)\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}\:\frac{\mathrm{du}}{\alpha} \\ $$$$=\int_{\mathrm{0}} ^{\alpha\mathrm{x}} \:\frac{\mathrm{udu}}{\left(\mathrm{u}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} \right)\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\:\mathrm{this}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{solvable}\:\mathrm{by}\:\mathrm{decomposition} \\ $$$$...\mathrm{be}\:\mathrm{continued}.... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com