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xdx-sin-2-x-3-

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

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