Question Number 201519 by vahid last updated on 08/Dec/23
Answered by cortano12 last updated on 08/Dec/23
$$\left(\mathrm{1}\right)\:\int\:\frac{\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}}\:\mathrm{dx}\:=\:−\int\:\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}}\:\mathrm{d}\left(\mathrm{cos}\:\mathrm{x}\right) \\ $$
Answered by Calculusboy last updated on 08/Dec/23
![2) I=∫(√((1−x)/(1+x))) dx let x=cos2𝛉 dx=−2sin2𝛉d𝛉 I=∫(√((1−cos2𝛉)/(1+cos2𝛉))) −2sin2𝛉d𝛉 Nb: 1−cos2𝛉=2sin^2 𝛉 and 1+cos2𝛉=2cos^2 𝛉 I=∫(√((2sin^2 𝛉)/(2cos^2 𝛉))) ∙−2sin2𝛉d𝛉 [Nb: sin2𝛉=2sin𝛉cos𝛉 and sin^2 𝛉=((1−cos2𝛉)/2)] I=−2∫((sin𝛉)/(cos𝛉))∙2sin𝛉cos𝛉d𝛉 ⇔ I=−4∫sin^2 𝛉d𝛉 I=−4∫(((1−cos2𝛉)/2))d𝛉 ⇔ I=−2∫(1−cos2𝛉)d𝛉 I=−2[∫1d𝛉−∫cos2𝛉d𝛉] I=−2[𝛉−((sin2𝛉)/2)]+C I=−2𝛉+sin𝛉+C but 𝛉=(1/2)cos^(−1) x I=cos^(−1) (x)+Sin[(1/2)cos^(−1) (x)]+C](https://www.tinkutara.com/question/Q201524.png)
$$\left.\mathrm{2}\right)\:\boldsymbol{{I}}=\int\sqrt{\frac{\mathrm{1}−\boldsymbol{{x}}}{\mathrm{1}+\boldsymbol{{x}}}}\:\boldsymbol{{dx}}\:\:\boldsymbol{{let}}\:\boldsymbol{{x}}=\boldsymbol{{cos}}\mathrm{2}\boldsymbol{\theta}\:\:\:\boldsymbol{{dx}}=−\mathrm{2}\boldsymbol{{sin}}\mathrm{2}\boldsymbol{\theta{d}\theta} \\ $$$$\boldsymbol{{I}}=\int\sqrt{\frac{\mathrm{1}−\boldsymbol{{cos}}\mathrm{2}\boldsymbol{\theta}}{\mathrm{1}+\boldsymbol{{cos}}\mathrm{2}\boldsymbol{\theta}}}\:−\mathrm{2}\boldsymbol{{sin}}\mathrm{2}\boldsymbol{\theta{d}\theta} \\ $$$$\boldsymbol{{Nb}}:\:\mathrm{1}−\boldsymbol{{cos}}\mathrm{2}\boldsymbol{\theta}=\mathrm{2}\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\theta}\:\:\boldsymbol{{and}}\:\:\mathrm{1}+\boldsymbol{{cos}}\mathrm{2}\boldsymbol{\theta}=\mathrm{2}\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{\theta} \\ $$$$\boldsymbol{{I}}=\int\sqrt{\frac{\mathrm{2}\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\theta}}{\mathrm{2}\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{\theta}}}\:\centerdot−\mathrm{2}\boldsymbol{{sin}}\mathrm{2}\boldsymbol{\theta{d}\theta}\:\:\:\left[\boldsymbol{{Nb}}:\:\boldsymbol{{sin}}\mathrm{2}\boldsymbol{\theta}=\mathrm{2}\boldsymbol{{sin}\theta{cos}\theta}\:\:\boldsymbol{{and}}\:\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\theta}=\frac{\mathrm{1}−\boldsymbol{{cos}}\mathrm{2}\boldsymbol{\theta}}{\mathrm{2}}\right] \\ $$$$\boldsymbol{{I}}=−\mathrm{2}\int\frac{\boldsymbol{{sin}\theta}}{\boldsymbol{{cos}\theta}}\centerdot\mathrm{2}\boldsymbol{{sin}\theta{cos}\theta{d}\theta}\:\:\:\Leftrightarrow\:\boldsymbol{{I}}=−\mathrm{4}\int\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\theta{d}\theta} \\ $$$$\boldsymbol{{I}}=−\mathrm{4}\int\left(\frac{\mathrm{1}−\boldsymbol{{cos}}\mathrm{2}\boldsymbol{\theta}}{\mathrm{2}}\right)\boldsymbol{{d}\theta}\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=−\mathrm{2}\int\left(\mathrm{1}−\boldsymbol{{cos}}\mathrm{2}\boldsymbol{\theta}\right)\boldsymbol{{d}\theta} \\ $$$$\boldsymbol{{I}}=−\mathrm{2}\left[\int\mathrm{1}\boldsymbol{{d}\theta}−\int\boldsymbol{{cos}}\mathrm{2}\boldsymbol{\theta{d}\theta}\right] \\ $$$$\boldsymbol{{I}}=−\mathrm{2}\left[\boldsymbol{\theta}−\frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{\theta}}{\mathrm{2}}\right]+\boldsymbol{{C}} \\ $$$$\boldsymbol{{I}}=−\mathrm{2}\boldsymbol{\theta}+\boldsymbol{{sin}\theta}+\boldsymbol{{C}}\:\:\:\:\boldsymbol{{but}}\:\boldsymbol{\theta}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{cos}}^{−\mathrm{1}} \boldsymbol{{x}} \\ $$$$\boldsymbol{{I}}=\boldsymbol{{cos}}^{−\mathrm{1}} \left(\boldsymbol{{x}}\right)+\boldsymbol{{Sin}}\left[\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{cos}}^{−\mathrm{1}} \left(\boldsymbol{{x}}\right)\right]+\boldsymbol{{C}} \\ $$$$ \\ $$
Answered by Calculusboy last updated on 08/Dec/23
![I=∫((sin^3 x)/(cos^6 x))dx Nb: sin^2 x=1−cos^2 x I=∫((sin^2 x ∙ sinx)/(cos^6 x))dx let u=cosx du=−sinxdx I=−∫(((1−cos^2 x)∙sinx)/u^6 )∙(du/(sinx)) I=−∫(((1−u^2 ))/u^6 )du ⇔ I=−∫[(1/u^6 )−(u^2 /u^6 )]du I=−((u^(−5) /(−5))−(u^(−3) /(−3)))+C I=(1/(5u^5 ))−(1/(3u^3 ))+C but u=cosx I=(1/(5cos^5 x))−(1/(3cos^3 x))+C](https://www.tinkutara.com/question/Q201525.png)
$$\boldsymbol{{I}}=\int\frac{\boldsymbol{{sin}}^{\mathrm{3}} \boldsymbol{{x}}}{\boldsymbol{{cos}}^{\mathrm{6}} \boldsymbol{{x}}}\boldsymbol{{dx}}\:\:\:\boldsymbol{{Nb}}:\:\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{{x}}=\mathrm{1}−\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}} \\ $$$$\boldsymbol{{I}}=\int\frac{\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{{x}}\:\centerdot\:\boldsymbol{{sinx}}}{\boldsymbol{{cos}}^{\mathrm{6}} \boldsymbol{{x}}}\boldsymbol{{dx}}\:\:\:\:\boldsymbol{{let}}\:\boldsymbol{{u}}=\boldsymbol{{cosx}}\:\:\boldsymbol{{du}}=−\boldsymbol{{sinxdx}} \\ $$$$\boldsymbol{{I}}=−\int\frac{\left(\mathrm{1}−\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}\right)\centerdot\boldsymbol{{sinx}}}{\boldsymbol{{u}}^{\mathrm{6}} }\centerdot\frac{\boldsymbol{{du}}}{\boldsymbol{{sinx}}} \\ $$$$\boldsymbol{{I}}=−\int\frac{\left(\mathrm{1}−\boldsymbol{{u}}^{\mathrm{2}} \right)}{\boldsymbol{{u}}^{\mathrm{6}} }\boldsymbol{{du}}\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=−\int\left[\frac{\mathrm{1}}{\boldsymbol{{u}}^{\mathrm{6}} }−\frac{\boldsymbol{{u}}^{\mathrm{2}} }{\boldsymbol{{u}}^{\mathrm{6}} }\right]\boldsymbol{{du}} \\ $$$$\boldsymbol{{I}}=−\left(\frac{\boldsymbol{{u}}^{−\mathrm{5}} }{−\mathrm{5}}−\frac{\boldsymbol{{u}}^{−\mathrm{3}} }{−\mathrm{3}}\right)+\boldsymbol{{C}} \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{1}}{\mathrm{5}\boldsymbol{{u}}^{\mathrm{5}} }−\frac{\mathrm{1}}{\mathrm{3}\boldsymbol{{u}}^{\mathrm{3}} }+\boldsymbol{{C}}\:\:\:\boldsymbol{{but}}\:\boldsymbol{{u}}=\boldsymbol{{cosx}} \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{1}}{\mathrm{5}\boldsymbol{{cos}}^{\mathrm{5}} \boldsymbol{{x}}}−\frac{\mathrm{1}}{\mathrm{3}\boldsymbol{{cos}}^{\mathrm{3}} \boldsymbol{{x}}}+\boldsymbol{{C}} \\ $$