Question Number 209709 by Ismoiljon_008 last updated on 19/Jul/24
$$ \\ $$$$\:\:\:\int\left({sinx}+{cosx}\right)^{\mathrm{11}} {dx}=\:? \\ $$$$\:\:\:{help}\:{me}\:{please} \\ $$
Answered by mr W last updated on 19/Jul/24
![∫(sin x+cos x)^(11) dx =∫((√2) cos (x−(π/4)))^(11) dx =32(√2)∫cos^(11) (x−(π/4))d(x−(π/4)) =32(√2)∫cos^(11) udu =32(√2)∫cos^(10) u d(sin u) =32(√2)∫(1−sin^2 u)^5 d(sin u) =32(√2)∫(1−t^2 )^5 dt =32(√2)∫(1−5t^2 +10t^4 −10t^6 +5t^8 −t^(10) )dt =32(√2)(t−((5t^3 )/3)+2t^5 −((10t^7 )/7)+((5t^9 )/9)−(t^(11) /(11)))+C =32(√2)[sin (x−(π/4))−((5 sin^3 (x−(π/4)))/3)+2 sin^5 (x−(π/4))−((10 sin^7 (x−(π/4)))/7)+((5 sin^9 (x−(π/4)))/9)−((sin^(11) (x−(π/4)))/(11))]+C](https://www.tinkutara.com/question/Q209710.png)
$$\int\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{\mathrm{11}} {dx} \\ $$$$=\int\left(\sqrt{\mathrm{2}}\:\mathrm{cos}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)\right)^{\mathrm{11}} {dx} \\ $$$$=\mathrm{32}\sqrt{\mathrm{2}}\int\mathrm{cos}^{\mathrm{11}} \:\left({x}−\frac{\pi}{\mathrm{4}}\right){d}\left({x}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\mathrm{32}\sqrt{\mathrm{2}}\int\mathrm{cos}^{\mathrm{11}} \:{udu} \\ $$$$=\mathrm{32}\sqrt{\mathrm{2}}\int\mathrm{cos}^{\mathrm{10}} \:{u}\:{d}\left(\mathrm{sin}\:{u}\right) \\ $$$$=\mathrm{32}\sqrt{\mathrm{2}}\int\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{u}\right)^{\mathrm{5}} {d}\left(\mathrm{sin}\:{u}\right) \\ $$$$=\mathrm{32}\sqrt{\mathrm{2}}\int\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{5}} {dt} \\ $$$$=\mathrm{32}\sqrt{\mathrm{2}}\int\left(\mathrm{1}−\mathrm{5}{t}^{\mathrm{2}} +\mathrm{10}{t}^{\mathrm{4}} −\mathrm{10}{t}^{\mathrm{6}} +\mathrm{5}{t}^{\mathrm{8}} −{t}^{\mathrm{10}} \right){dt} \\ $$$$=\mathrm{32}\sqrt{\mathrm{2}}\left({t}−\frac{\mathrm{5}{t}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2}{t}^{\mathrm{5}} −\frac{\mathrm{10}{t}^{\mathrm{7}} }{\mathrm{7}}+\frac{\mathrm{5}{t}^{\mathrm{9}} }{\mathrm{9}}−\frac{{t}^{\mathrm{11}} }{\mathrm{11}}\right)+{C} \\ $$$$=\mathrm{32}\sqrt{\mathrm{2}}\left[\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)−\frac{\mathrm{5}\:\mathrm{sin}^{\mathrm{3}} \:\left({x}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{3}}+\mathrm{2}\:\mathrm{sin}^{\mathrm{5}} \:\left({x}−\frac{\pi}{\mathrm{4}}\right)−\frac{\mathrm{10}\:\mathrm{sin}^{\mathrm{7}} \:\left({x}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{7}}+\frac{\mathrm{5}\:\mathrm{sin}^{\mathrm{9}} \:\left({x}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{9}}−\frac{\mathrm{sin}^{\mathrm{11}} \:\left({x}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{11}}\right]+{C} \\ $$
Commented by Ismoiljon_008 last updated on 19/Jul/24
$$\:\:\:{thank}\:{you}\:{very}\:{much}\: \\ $$