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Question Number 112699 by bemath last updated on 09/Sep/20

(1) ∫ ((sin x e^(√(cos x)) )/( (√(cos x)))) dx (2) ((1+(√(1+sin 2A)))/(1−(√(1+sin 2A))))?? (3)∫ (dx/((x+b)(x^2 +a^2 )))

$$\:\left(\mathrm{1}\right)\:\int\:\frac{\mathrm{sin}\:\mathrm{x}\:\mathrm{e}^{\sqrt{\mathrm{cos}\:\mathrm{x}}} }{\:\sqrt{\mathrm{cos}\:\mathrm{x}}}\:\mathrm{dx}\: \\ $$$$\:\left(\mathrm{2}\right)\:\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{2A}}}{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{2A}}}??\: \\ $$$$\left(\mathrm{3}\right)\int\:\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{b}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)} \\ $$

Commented by Dwaipayan Shikari last updated on 09/Sep/20

∫((sinxe^(√(cosx)) )/( (√(cosx))))dx cosx=t^2 , −sinx=2t(dt/dx) −∫((2te^t )/t)dt=−2e^t +C=−2e^(√(cosx)) +C

$$\int\frac{{sinxe}^{\sqrt{{cosx}}} }{\:\sqrt{{cosx}}}{dx}\:\:\:\:\:\:\:\:\:\:\:{cosx}={t}^{\mathrm{2}} ,\:−{sinx}=\mathrm{2}{t}\frac{{dt}}{{dx}} \\ $$$$−\int\frac{\mathrm{2}{te}^{{t}} }{{t}}{dt}=−\mathrm{2}{e}^{{t}} +{C}=−\mathrm{2}{e}^{\sqrt{{cosx}}} +{C} \\ $$

Answered by bobhans last updated on 09/Sep/20

(⧫) ∫ ((sin x e^(√(cos x)) )/( (√(cos x)))) dx=I let u = (√(cos x)) →du = ((−sin x)/(2(√(cos x)))) dx I = ∫ e^u (−2du) = −2e^u + c I=−2e^((√(cos x)) ) + c

$$\left(\blacklozenge\right)\:\int\:\frac{\mathrm{sin}\:\mathrm{x}\:\mathrm{e}^{\sqrt{\mathrm{cos}\:\mathrm{x}}} }{\:\sqrt{\mathrm{cos}\:\mathrm{x}}}\:\mathrm{dx}=\mathrm{I} \\ $$$$\mathrm{let}\:\mathrm{u}\:=\:\sqrt{\mathrm{cos}\:\mathrm{x}}\:\rightarrow\mathrm{du}\:=\:\frac{−\mathrm{sin}\:\mathrm{x}}{\mathrm{2}\sqrt{\mathrm{cos}\:\mathrm{x}}}\:\mathrm{dx} \\ $$$$\mathrm{I}\:=\:\int\:\mathrm{e}^{\mathrm{u}} \left(−\mathrm{2du}\right)\:=\:−\mathrm{2e}^{\mathrm{u}} \:+\:\mathrm{c} \\ $$$$\mathrm{I}=−\mathrm{2e}^{\sqrt{\mathrm{cos}\:\mathrm{x}}\:} \:+\:\mathrm{c}\: \\ $$

Answered by bobhans last updated on 09/Sep/20

(⧫⧫) ((1+(√(1+sin 2A)))/(1−(√(1+sin 2A)))) = ((1+sin A+cos A)/(1−(sin A+cos A))) = ((1+2sin ((A/2))cos ((A/2))+2cos^2 ((A/2))−1)/(1−2sin ((A/2))cos ((A/2))−(1−2sin^2 ((A/2))))) = ((2cos ((A/2))(sin ((A/2))+cos ((A/2))))/(2sin ((A/2))( sin ((A/2))−cos ((A/2))))) = cot ((A/2)).((tan ((A/2))+1)/(tan ((A/2))−1))

$$\left(\blacklozenge\blacklozenge\right)\:\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{2A}}}{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{2A}}}\:=\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{A}+\mathrm{cos}\:\mathrm{A}}{\mathrm{1}−\left(\mathrm{sin}\:\mathrm{A}+\mathrm{cos}\:\mathrm{A}\right)} \\ $$$$\:=\:\frac{\mathrm{1}+\mathrm{2sin}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)+\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)−\mathrm{1}}{\mathrm{1}−\mathrm{2sin}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)\right)} \\ $$$$=\:\frac{\mathrm{2cos}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)\left(\mathrm{sin}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)\right)}{\mathrm{2sin}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)\left(\:\mathrm{sin}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)−\mathrm{cos}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)\right)} \\ $$$$=\:\mathrm{cot}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right).\frac{\mathrm{tan}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)+\mathrm{1}}{\mathrm{tan}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)−\mathrm{1}} \\ $$$$ \\ $$

Answered by 1549442205PVT last updated on 09/Sep/20

P=((1+(√(1+sin2A)))/(1−(√(1+sin2A))))=((1+(√((sinA+cosA)^2 )))/(1−(√((sinA+cosA))))) =((1+∣sinA+cosA∣)/(1−∣sinA+cosA∣)) i)If sinA+cosA≥0 then P=(((1+cosA)+sinA)/((1+cosA)−sinA))=((2cos^2 (A/2)+2sin(A/2)cos(A/2))/(2cos^2 (A/2)−2sin(A/2)cos(A/2))) =((2cos(A/2)(cos(A/2)+sin(A/2)))/(2cos(A/2)(cos(A/2)−sin(A/2))))=((cos(A/2)+sin(A/2))/(cos(A/2)−sin(A/2))) =(((√2) cos((𝛑/4)−(A/2)))/( (√2) cos((𝛑/4)+(A/2))))=((cos((𝛑/4)−(A/2)))/(cos((𝛑/4)+(A/2)))) ii)If sinA+cosA<0 then .Similarly, we get P=((1−sinA−cosA)/(1+sinA+cosA)) =((2sin^2 (A/2)−2sin(A/2)cos(A/2))/(2cos^2 (A/2)+2sin(A/2)cos(A/2)))=((2sin(A/2)(sin(A/2)−cos(A/2)))/(2cos(A/2)(cos(A/2)+sin(A/2)))) =tan(A/2)×((sin((𝛑/4)−(A/2)))/(sin((𝛑/4)+(A/2)))) 3)Find ∫ (dx/((x+b)(x^2 +a^2 ))) F=∫ (dx/((x+b)(x^2 +a^2 )))=∫(dx/((a^2 +b^2 )(x+b))) −∫ ((x−b)/((a^2 +b^2 )(x^2 +a^2 )))dx =(1/(a^2 +b^2 ))ln∣x+b∣−(1/(a^2 +b^2 ))×(1/(2a))tan^(−1) ((x/a))

$$\mathrm{P}=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{sin2A}}}{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{sin2A}}}=\frac{\mathrm{1}+\sqrt{\left(\mathrm{sinA}+\mathrm{cosA}\right)^{\mathrm{2}} }}{\mathrm{1}−\sqrt{\left(\mathrm{sinA}+\mathrm{cosA}\right)}} \\ $$$$=\frac{\mathrm{1}+\mid\mathrm{sinA}+\mathrm{cosA}\mid}{\mathrm{1}−\mid\mathrm{sinA}+\mathrm{cosA}\mid} \\ $$$$\left.\mathrm{i}\right)\mathrm{If}\:\mathrm{sinA}+\mathrm{cosA}\geqslant\mathrm{0}\:\mathrm{then} \\ $$$$\mathrm{P}=\frac{\left(\mathrm{1}+\mathrm{cosA}\right)+\mathrm{sinA}}{\left(\mathrm{1}+\mathrm{cosA}\right)−\mathrm{sinA}}=\frac{\mathrm{2cos}^{\mathrm{2}} \frac{\mathrm{A}}{\mathrm{2}}+\mathrm{2sin}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{A}}{\mathrm{2}}}{\mathrm{2cos}^{\mathrm{2}} \frac{\mathrm{A}}{\mathrm{2}}−\mathrm{2sin}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{A}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{2cos}\frac{\mathrm{A}}{\mathrm{2}}\left(\mathrm{cos}\frac{\mathrm{A}}{\mathrm{2}}+\mathrm{sin}\frac{\mathrm{A}}{\mathrm{2}}\right)}{\mathrm{2cos}\frac{\mathrm{A}}{\mathrm{2}}\left(\mathrm{cos}\frac{\mathrm{A}}{\mathrm{2}}−\mathrm{sin}\frac{\mathrm{A}}{\mathrm{2}}\right)}=\frac{\mathrm{cos}\frac{\mathrm{A}}{\mathrm{2}}+\mathrm{sin}\frac{\mathrm{A}}{\mathrm{2}}}{\mathrm{cos}\frac{\mathrm{A}}{\mathrm{2}}−\mathrm{sin}\frac{\mathrm{A}}{\mathrm{2}}} \\ $$$$=\frac{\sqrt{\mathrm{2}}\:\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}−\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{2}}\:\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}+\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}\right)}=\frac{\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}−\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}\right)}{\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}+\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}\right)}\: \\ $$$$\left.\mathrm{ii}\right)\mathrm{If}\:\mathrm{sinA}+\mathrm{cosA}<\mathrm{0}\:\mathrm{then}\:.\mathrm{Similarly}, \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{P}=\frac{\mathrm{1}−\mathrm{sinA}−\mathrm{cosA}}{\mathrm{1}+\mathrm{sinA}+\mathrm{cosA}} \\ $$$$=\frac{\mathrm{2sin}^{\mathrm{2}} \frac{\mathrm{A}}{\mathrm{2}}−\mathrm{2sin}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{A}}{\mathrm{2}}}{\mathrm{2cos}^{\mathrm{2}} \frac{\mathrm{A}}{\mathrm{2}}+\mathrm{2sin}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{A}}{\mathrm{2}}}=\frac{\mathrm{2sin}\frac{\mathrm{A}}{\mathrm{2}}\left(\mathrm{sin}\frac{\mathrm{A}}{\mathrm{2}}−\mathrm{cos}\frac{\mathrm{A}}{\mathrm{2}}\right)}{\mathrm{2cos}\frac{\mathrm{A}}{\mathrm{2}}\left(\mathrm{cos}\frac{\mathrm{A}}{\mathrm{2}}+\mathrm{sin}\frac{\mathrm{A}}{\mathrm{2}}\right)} \\ $$$$=\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}×\frac{\boldsymbol{\mathrm{sin}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}−\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}\right)}{\boldsymbol{\mathrm{sin}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}+\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}\right)}\: \\ $$$$\left.\mathrm{3}\right)\boldsymbol{\mathrm{Find}}\:\int\:\frac{\boldsymbol{\mathrm{dx}}}{\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{b}}\right)\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)} \\ $$$$\mathrm{F}=\int\:\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{b}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)}=\int\frac{\mathrm{dx}}{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)\left(\mathrm{x}+\mathrm{b}\right)} \\ $$$$−\int\:\frac{\mathrm{x}−\mathrm{b}}{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} }\boldsymbol{\mathrm{ln}}\mid\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{b}}\mid−\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{a}}}\boldsymbol{{tan}}^{−\mathrm{1}} \left(\frac{\boldsymbol{{x}}}{\boldsymbol{{a}}}\right) \\ $$

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