Question-109457

Question Number 109457 by shahria14 last updated on 23/Aug/20

Answered by Dwaipayan Shikari last updated on 23/Aug/20

∫sin^3 x(1−cos^2 x)dx ∫sin^3 x−∫sin^3 xcos^2 x ∫sinx(1−cos^2 x)+∫sin^2 x(−sinx)cos^2 xdx ∫sinx−∫sinxcos^2 x+∫(1−t^2 )t^2 dt −cosx+((cos^3 x)/3)+((cos^3 x)/3)−((cos^5 x)/5)+C

$$\int{sin}^{\mathrm{3}} {x}\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right){dx} \\ $$$$\int{sin}^{\mathrm{3}} {x}−\int{sin}^{\mathrm{3}} {xcos}^{\mathrm{2}} {x} \\ $$$$\int{sinx}\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)+\int{sin}^{\mathrm{2}} {x}\left(−{sinx}\right){cos}^{\mathrm{2}} {xdx} \\ $$$$\int{sinx}−\int{sinxcos}^{\mathrm{2}} {x}+\int\left(\mathrm{1}−{t}^{\mathrm{2}} \right){t}^{\mathrm{2}} {dt} \\ $$$$−{cosx}+\frac{{cos}^{\mathrm{3}} {x}}{\mathrm{3}}+\frac{{cos}^{\mathrm{3}} {x}}{\mathrm{3}}−\frac{{cos}^{\mathrm{5}} {x}}{\mathrm{5}}+{C} \\ $$

Answered by john santu last updated on 24/Aug/20

((✓JS✓)/(“•“)) ∫ sin^4 x sin x dx = ∫ (1−cos^2 x)^2 (−d(cos x)) ∫(1−n^2 )^2 (−dn) = −∫[ n^4 −2n^2 +1] dn =−[ (1/5)n^5 −(2/3)n^3 +n ] + c =−((cos^5 x)/5)+((2cos^3 x)/3)−cos x + c

$$\:\:\:\:\:\:\frac{\checkmark{JS}\checkmark}{“\bullet“} \\ $$$$\:\:\:\int\:\mathrm{sin}\:^{\mathrm{4}} {x}\:\mathrm{sin}\:{x}\:{dx}\:=\:\int\:\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} \left(−{d}\left(\mathrm{cos}\:{x}\right)\right) \\ $$$$\int\left(\mathrm{1}−{n}^{\mathrm{2}} \right)^{\mathrm{2}} \left(−{dn}\right)\:=\:−\int\left[\:{n}^{\mathrm{4}} −\mathrm{2}{n}^{\mathrm{2}} +\mathrm{1}\right]\:{dn} \\ $$$$=−\left[\:\frac{\mathrm{1}}{\mathrm{5}}{n}^{\mathrm{5}} −\frac{\mathrm{2}}{\mathrm{3}}{n}^{\mathrm{3}} +{n}\:\right]\:+\:{c} \\ $$$$=−\frac{\mathrm{cos}\:^{\mathrm{5}} {x}}{\mathrm{5}}+\frac{\mathrm{2cos}\:^{\mathrm{3}} {x}}{\mathrm{3}}−\mathrm{cos}\:{x}\:+\:{c}\: \\ $$

Answered by 1549442205PVT last updated on 24/Aug/20

∫sin^5 xdx=−∫sin^4 xd(cosx)=−∫(1−cos^2 x)^2 d(cosx) =∫(−1+2cos^2 x−cos^4 x)d(cosx) =−cosx+((2cos^3 x)/3)−((cos^5 x)/5)+C

$$\int\mathrm{sin}^{\mathrm{5}} \mathrm{xdx}=−\int\mathrm{sin}^{\mathrm{4}} \mathrm{xd}\left(\mathrm{cosx}\right)=−\int\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{2}} \mathrm{d}\left(\mathrm{cosx}\right) \\ $$$$=\int\left(−\mathrm{1}+\mathrm{2cos}^{\mathrm{2}} \mathrm{x}−\mathrm{cos}^{\mathrm{4}} \mathrm{x}\right)\mathrm{d}\left(\mathrm{cosx}\right) \\ $$$$=−\boldsymbol{\mathrm{cosx}}+\frac{\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{3}} \boldsymbol{\mathrm{x}}}{\mathrm{3}}−\frac{\boldsymbol{\mathrm{cos}}^{\mathrm{5}} \boldsymbol{\mathrm{x}}}{\mathrm{5}}+\boldsymbol{\mathrm{C}} \\ $$

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