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Question Number 84163 by john santu last updated on 10/Mar/20
∫ sin (50x) sin^(49) (x) dx ?
$$\int\:\mathrm{sin}\:\left(\mathrm{50}{x}\right)\:\mathrm{sin}\:^{\mathrm{49}} \left({x}\right)\:{dx}\:? \\ $$
Answered by jagoll last updated on 10/Mar/20
sin 50x = sin 49x cos x + cos 49x sin x ∫ sin^(49) x sin 49x cos x dx + ∫ cos 49x sin^(50 ) x dx let u = sin^(50) x ⇒ du = 50 sin^(49) x cos x dx v = sin 49x ⇒ dv = 49 cos 49x dx ⇒ sin^(49) x cos x sin 49x = (1/(50))v (du/dx) ; ⇒ sin^(50) x cos 49x = (1/(50))u (dv/dx) then ∫ ((1/(50)) v (du/dx) + (1/(50)) u(dv/dx) ) dx = ∫ (1/(50)) ((d(u.v))/dx) × dx = (1/(50)) u.v + c = (1/(50)) sin(49x) sin^(50) (x) + c
$$\mathrm{sin}\:\mathrm{50x}\:=\:\mathrm{sin}\:\mathrm{49x}\:\mathrm{cos}\:\mathrm{x}\:+\:\mathrm{cos}\:\mathrm{49x}\:\mathrm{sin}\:\mathrm{x} \\ $$$$\int\:\mathrm{sin}\:^{\mathrm{49}} \mathrm{x}\:\mathrm{sin}\:\mathrm{49x}\:\mathrm{cos}\:\mathrm{x}\:\mathrm{dx}\:+ \\ $$$$\int\:\mathrm{cos}\:\mathrm{49x}\:\mathrm{sin}^{\mathrm{50}\:} \mathrm{x}\:\mathrm{dx} \\ $$$$\mathrm{let}\:\mathrm{u}\:=\:\mathrm{sin}\:^{\mathrm{50}} \mathrm{x}\:\Rightarrow\:\mathrm{du}\:=\:\mathrm{50}\:\mathrm{sin}\:^{\mathrm{49}} \mathrm{x}\:\mathrm{cos}\:\mathrm{x}\:\mathrm{dx} \\ $$$$\mathrm{v}\:=\:\mathrm{sin}\:\mathrm{49x}\:\Rightarrow\:\mathrm{dv}\:=\:\mathrm{49}\:\mathrm{cos}\:\mathrm{49x}\:\mathrm{dx} \\ $$$$\Rightarrow\:\mathrm{sin}\:^{\mathrm{49}} \mathrm{x}\:\mathrm{cos}\:\mathrm{x}\:\mathrm{sin}\:\mathrm{49x}\:=\:\frac{\mathrm{1}}{\mathrm{50}}\mathrm{v}\:\frac{\mathrm{du}}{\mathrm{dx}}\:;\: \\ $$$$\Rightarrow\:\mathrm{sin}\:^{\mathrm{50}} \mathrm{x}\:\mathrm{cos}\:\mathrm{49x}\:=\:\frac{\mathrm{1}}{\mathrm{50}}\mathrm{u}\:\frac{\mathrm{dv}}{\mathrm{dx}} \\ $$$$\mathrm{then}\: \\ $$$$\int\:\left(\frac{\mathrm{1}}{\mathrm{50}}\:\mathrm{v}\:\frac{\mathrm{du}}{\mathrm{dx}}\:+\:\frac{\mathrm{1}}{\mathrm{50}}\:\mathrm{u}\frac{\mathrm{dv}}{\mathrm{dx}}\:\right)\:\mathrm{dx}\:=\: \\ $$$$\int\:\frac{\mathrm{1}}{\mathrm{50}}\:\frac{\mathrm{d}\left(\mathrm{u}.\mathrm{v}\right)}{\mathrm{dx}}\:×\:\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{50}}\:\mathrm{u}.\mathrm{v}\:+\:\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{50}}\:\mathrm{sin}\left(\mathrm{49x}\right)\:\mathrm{sin}\:^{\mathrm{50}} \:\left(\mathrm{x}\right)\:+\:\mathrm{c}\: \\ $$$$ \\ $$
Commented by john santu last updated on 10/Mar/20
good sir
$${good}\:{sir} \\ $$
Commented by TANMAY PANACEA last updated on 10/Mar/20
pls check...(dv/dx)=49cos49x so cos49x=(1/(49))(dv/dx) so sin^(50) xcos49x=u×(1/(49))(dv/dx)≠(1/(50))u(dv/dx) pls check
$${pls}\:{check}…\frac{{dv}}{{dx}}=\mathrm{49}{cos}\mathrm{49}{x}\:\:{so}\:\boldsymbol{{cos}}\mathrm{49}\boldsymbol{{x}}=\frac{\mathrm{1}}{\mathrm{49}}\frac{\boldsymbol{{dv}}}{\boldsymbol{{dx}}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{sin}}^{\mathrm{50}} \boldsymbol{{xcos}}\mathrm{49}\boldsymbol{{x}}=\boldsymbol{{u}}×\frac{\mathrm{1}}{\mathrm{49}}\frac{\boldsymbol{{dv}}}{\boldsymbol{{dx}}}\neq\frac{\mathrm{1}}{\mathrm{50}}{u}\frac{{dv}}{{dx}} \\ $$$${pls}\:{check} \\ $$
Answered by TANMAY PANACEA last updated on 10/Mar/20
∫(sin49x.cosx.sin^(49) x+co49xsin^(50) x)dx I_1 =∫sin49x.cosx.sin^(49) xdx I_2 =∫cos49xsin^(50) xdx I=I_1 +I_2 I_1 =sin49x∫sin^(49) xcosxdx−∫[((d sin49x)/dx)∫sin^(49) xcosxdx]dx =sin49x.((sin^(50) x)/(50))−∫49cos49x.((sin^(50) x)/(50))dx =((sin49x.sin^(50) x)/(50))−((49)/(50))I_2 I=I_1 +I_2 =((sin49x.sin^(50) x)/(50))−((49)/(50))I_2 +I_2 I=((sin49x.sin^(50) x)/(50))+(I_2 /(50)) wait pls
$$\int\left({sin}\mathrm{49}{x}.{cosx}.{sin}^{\mathrm{49}} {x}+{co}\mathrm{49}{xsin}^{\mathrm{50}} {x}\right){dx} \\ $$$${I}_{\mathrm{1}} =\int{sin}\mathrm{49}{x}.{cosx}.{sin}^{\mathrm{49}} {xdx} \\ $$$${I}_{\mathrm{2}} =\int{cos}\mathrm{49}{xsin}^{\mathrm{50}} {xdx} \\ $$$${I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} ={sin}\mathrm{49}{x}\int{sin}^{\mathrm{49}} {xcosxdx}−\int\left[\frac{{d}\:{sin}\mathrm{49}{x}}{{dx}}\int{sin}^{\mathrm{49}} {xcosxdx}\right]{dx} \\ $$$$={sin}\mathrm{49}{x}.\frac{{sin}^{\mathrm{50}} {x}}{\mathrm{50}}−\int\mathrm{49}{cos}\mathrm{49}{x}.\frac{{sin}^{\mathrm{50}} {x}}{\mathrm{50}}{dx} \\ $$$$=\frac{{sin}\mathrm{49}{x}.{sin}^{\mathrm{50}} {x}}{\mathrm{50}}−\frac{\mathrm{49}}{\mathrm{50}}{I}_{\mathrm{2}} \\ $$$${I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$$\:\:\:=\frac{{sin}\mathrm{49}{x}.{sin}^{\mathrm{50}} {x}}{\mathrm{50}}−\frac{\mathrm{49}}{\mathrm{50}}{I}_{\mathrm{2}} +{I}_{\mathrm{2}} \\ $$$${I}=\frac{{sin}\mathrm{49}{x}.{sin}^{\mathrm{50}} {x}}{\mathrm{50}}+\frac{{I}_{\mathrm{2}} }{\mathrm{50}} \\ $$$${wait}\:{pls} \\ $$$$ \\ $$

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