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Question Number 108384 by bobhans last updated on 16/Aug/20

(1)∫ ((√(sin x))/( (√(sin x)) + (√(cos x)))) dx ? (2) (d^2 y/dx^2 )−2(dy/dx) +y = e^x

$$\left(\mathrm{1}\right)\int\:\frac{\sqrt{\mathrm{sin}\:{x}}}{\:\sqrt{\mathrm{sin}\:{x}}\:+\:\sqrt{\mathrm{cos}\:{x}}}\:{dx}\:? \\ $$$$\left(\mathrm{2}\right)\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\mathrm{2}\frac{{dy}}{{dx}}\:+{y}\:=\:{e}^{{x}} \\ $$

Answered by bemath last updated on 16/Aug/20

Commented by bobhans last updated on 16/Aug/20

yes.....

$${yes}..... \\ $$

Answered by Sarah85 last updated on 16/Aug/20

∫((√(sin x))/( (√(sin x))+(√(cos x))))dx let t=(√(tan x)) ⇔ x=arctan t^2 ⇔ dx=((2t)/(t^4 +1))dt ∫((2t^2 )/((t+1)(t^4 +1)))dt= =∫(1/(t+1))dt−∫((t^3 −t^2 −t+1)/(t^4 +1))dt =∫(1/(t+1))dt−∫(((1−(√2))(t+1))/(2(t^2 −(√2)t+1)))dt−∫(((1+(√2))(t+1))/(2(t^2 +(√2)t+1)))dt now I use the obvious formulas to get ln ∣t+1∣ −((1−(√2))/4)ln (t^2 −(√2)t+1) +(1/2)arctan ((√2)t−1) −((1+(√2))/4)ln (t^2 +(√2)t+1) −(1/2)arctan ((√2)t+1) now insert t=(√(tan x))

$$\int\frac{\sqrt{\mathrm{sin}\:{x}}}{\:\sqrt{\mathrm{sin}\:{x}}+\sqrt{\mathrm{cos}\:{x}}}{dx} \\ $$$$\mathrm{let}\:{t}=\sqrt{\mathrm{tan}\:{x}}\:\Leftrightarrow\:{x}=\mathrm{arctan}\:{t}^{\mathrm{2}} \:\Leftrightarrow\:{dx}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt} \\ $$$$\int\frac{\mathrm{2}{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{4}} +\mathrm{1}\right)}{dt}= \\ $$$$=\int\frac{\mathrm{1}}{{t}+\mathrm{1}}{dt}−\int\frac{{t}^{\mathrm{3}} −{t}^{\mathrm{2}} −{t}+\mathrm{1}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt} \\ $$$$=\int\frac{\mathrm{1}}{{t}+\mathrm{1}}{dt}−\int\frac{\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\left({t}+\mathrm{1}\right)}{\mathrm{2}\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)}{dt}−\int\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left({t}+\mathrm{1}\right)}{\mathrm{2}\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)}{dt} \\ $$$$\mathrm{now}\:\mathrm{I}\:\mathrm{use}\:\mathrm{the}\:\mathrm{obvious}\:\mathrm{formulas}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{ln}\:\mid{t}+\mathrm{1}\mid\:−\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right)\:−\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{t}+\mathrm{1}\right) \\ $$$$\mathrm{now}\:\mathrm{insert}\:{t}=\sqrt{\mathrm{tan}\:{x}} \\ $$

Answered by mathmax by abdo last updated on 17/Aug/20

2) y^(′′) −2y^′ +y =e^x h→r^2 −2r +1 =0 ⇒(r−1)^2 =0 ⇒r=1(double) ⇒y_h =(ax+b)e^x =axe^x +be^x =au_1 +bu_2 W(u_1 ,u_2 ) = determinant (((xe^x e^x )),(((x+1)e^x e^x )))=xe^(2x) −(x+1)e^(2x) =−e^(2x) ≠0 W_1 = determinant (((o e^x )),((e^x e^x )))=−e^(2x) W_2 = determinant (((xe^x 0)),(((x+1)e^x e^x )))=xe^(2x) V_1 =∫ (W_1 /W)dx =∫ ((−e^(2x) )/(−e^(2x) ))dx =x V_2 =∫ (W_2 /W)dx =∫ ((xe^(2x) )/(−e^(2x) ))dx =−(x^2 /2) ⇒ y_p =u_1 v_1 +u_2 v_2 =xe^x (x)+e^x (−(x^2 /2)) =(x^2 /2)e^x ⇒the general solution is y =y_h +y_p =(ax+b)e^x +(x^2 /2) e^x

$$\left.\mathrm{2}\right)\:\mathrm{y}^{''} −\mathrm{2y}^{'} \:+\mathrm{y}\:=\mathrm{e}^{\mathrm{x}} \: \\ $$$$\mathrm{h}\rightarrow\mathrm{r}^{\mathrm{2}} −\mathrm{2r}\:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow\left(\mathrm{r}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\mathrm{r}=\mathrm{1}\left(\mathrm{double}\right)\:\Rightarrow\mathrm{y}_{\mathrm{h}} =\left(\mathrm{ax}+\mathrm{b}\right)\mathrm{e}^{\mathrm{x}} \\ $$$$=\mathrm{axe}^{\mathrm{x}} \:+\mathrm{be}^{\mathrm{x}} \:=\mathrm{au}_{\mathrm{1}} \:+\mathrm{bu}_{\mathrm{2}} \\ $$$$\mathrm{W}\left(\mathrm{u}_{\mathrm{1}} ,\mathrm{u}_{\mathrm{2}} \right)\:=\begin{vmatrix}{\mathrm{xe}^{\mathrm{x}} \:\:\:\:\:\:\:\:\:\mathrm{e}^{\mathrm{x}} }\\{\left(\mathrm{x}+\mathrm{1}\right)\mathrm{e}^{\mathrm{x}} \:\:\:\mathrm{e}^{\mathrm{x}} }\end{vmatrix}=\mathrm{xe}^{\mathrm{2x}} −\left(\mathrm{x}+\mathrm{1}\right)\mathrm{e}^{\mathrm{2x}} \:=−\mathrm{e}^{\mathrm{2x}} \:\neq\mathrm{0} \\ $$$$\mathrm{W}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{o}\:\:\:\:\:\:\:\:\:\mathrm{e}^{\mathrm{x}} }\\{\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\:\mathrm{e}^{\mathrm{x}} }\end{vmatrix}=−\mathrm{e}^{\mathrm{2x}} \\ $$$$\mathrm{W}_{\mathrm{2}} =\begin{vmatrix}{\mathrm{xe}^{\mathrm{x}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\left(\mathrm{x}+\mathrm{1}\right)\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\mathrm{e}^{\mathrm{x}} }\end{vmatrix}=\mathrm{xe}^{\mathrm{2x}} \\ $$$$\mathrm{V}_{\mathrm{1}} =\int\:\frac{\mathrm{W}_{\mathrm{1}} }{\mathrm{W}}\mathrm{dx}\:=\int\:\:\frac{−\mathrm{e}^{\mathrm{2x}} }{−\mathrm{e}^{\mathrm{2x}} }\mathrm{dx}\:=\mathrm{x} \\ $$$$\mathrm{V}_{\mathrm{2}} =\int\:\frac{\mathrm{W}_{\mathrm{2}} }{\mathrm{W}}\mathrm{dx}\:=\int\:\:\frac{\mathrm{xe}^{\mathrm{2x}} }{−\mathrm{e}^{\mathrm{2x}} }\mathrm{dx}\:=−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{y}_{\mathrm{p}} =\mathrm{u}_{\mathrm{1}} \mathrm{v}_{\mathrm{1}} \:+\mathrm{u}_{\mathrm{2}} \mathrm{v}_{\mathrm{2}} =\mathrm{xe}^{\mathrm{x}} \left(\mathrm{x}\right)+\mathrm{e}^{\mathrm{x}} \left(−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)\:=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{e}^{\mathrm{x}} \:\Rightarrow\mathrm{the}\:\mathrm{general}\:\mathrm{solution} \\ $$$$\mathrm{is}\:\mathrm{y}\:=\mathrm{y}_{\mathrm{h}} \:+\mathrm{y}_{\mathrm{p}} =\left(\mathrm{ax}+\mathrm{b}\right)\mathrm{e}^{\mathrm{x}} \:+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\mathrm{e}^{\mathrm{x}} \\ $$

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