Question Number 146772 by ZiYangLee last updated on 15/Jul/21
$$\mathrm{If}\:{z}=\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta,\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{cos}^{\mathrm{6}} \theta=\frac{\mathrm{1}}{\mathrm{32}}\left(\mathrm{cos}\:\mathrm{6}\theta+\mathrm{6cos}\:\mathrm{4}\theta+\mathrm{15cos}\:\mathrm{2}\theta+\mathrm{10}\right). \\ $$$$\mathrm{Hence}\:\mathrm{or}\:\mathrm{otherwise},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:{a}} \sqrt{\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\mathrm{5}} }\:{dx}. \\ $$
Answered by mathmax by abdo last updated on 15/Jul/21
![Ψ=∫_0 ^a (a^2 −x^2 )^(5/2) dx changement x=asint give Ψ=∫_0 ^(π/2) a^5 (cos^5 θ)acosθ dθ =a^6 ∫_0 ^(π/2) cos^6 θ dθ =(a^6 /(32)){ ∫_0 ^(π/2) cos(6θ)dθ +6∫_0 ^(π/2) cos(4θ)dθ +15∫_0 ^(π/2) cos(2θ)dθ +10∫_0 ^(π/2) [dθ} =(a^6 /(32)){[(1/6)sin(6θ)]_0 ^(π/2) +6[(1/4)sin(4θ)]_0 ^(π/2) +15[(1/2)sin(2θ)]_0 ^(π/2) +5π} =(a^6 /(32)){0+0+0+5π}=((5π)/(32))a^6 cos^6 θ =(((e^(iθ) +e^(−iθ) )/2))^6 =(1/2^6 )Σ_(k=0) ^6 C_6 ^k (e^(iθ) )^k (e^(−iθ) )^(6−k) =(1/2^6 )Σ_(k=0) ^6 C_6 ^k e^(ikθ+ikθ−6iθ) =(1/2^6 )Σ_(k=0) ^6 C_6 ^k e^(i(2k−6)θ) =.....](Q146803.png)
$$\Psi=\int_{\mathrm{0}} ^{\mathrm{a}} \left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} \:\mathrm{dx}\:\mathrm{changement}\:\mathrm{x}=\mathrm{asint}\:\mathrm{give} \\ $$$$\Psi=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{a}^{\mathrm{5}} \left(\mathrm{cos}^{\mathrm{5}} \theta\right)\mathrm{acos}\theta\:\mathrm{d}\theta\:=\mathrm{a}^{\mathrm{6}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{6}} \:\theta\:\mathrm{d}\theta \\ $$$$=\frac{\mathrm{a}^{\mathrm{6}} }{\mathrm{32}}\left\{\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}\left(\mathrm{6}\theta\right)\mathrm{d}\theta\:+\mathrm{6}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}\left(\mathrm{4}\theta\right)\mathrm{d}\theta\:+\mathrm{15}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\left(\mathrm{2}\theta\right)\mathrm{d}\theta\:+\mathrm{10}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[\mathrm{d}\theta\right\}\right. \\ $$$$=\frac{\mathrm{a}^{\mathrm{6}} }{\mathrm{32}}\left\{\left[\frac{\mathrm{1}}{\mathrm{6}}\mathrm{sin}\left(\mathrm{6}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:+\mathrm{6}\left[\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{4}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\mathrm{15}\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{2}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:+\mathrm{5}\pi\right\} \\ $$$$=\frac{\mathrm{a}^{\mathrm{6}} }{\mathrm{32}}\left\{\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{5}\pi\right\}=\frac{\mathrm{5}\pi}{\mathrm{32}}\mathrm{a}^{\mathrm{6}} \\ $$$$\mathrm{cos}^{\mathrm{6}} \theta\:=\left(\frac{\mathrm{e}^{\mathrm{i}\theta} \:+\mathrm{e}^{−\mathrm{i}\theta} }{\mathrm{2}}\right)^{\mathrm{6}} \:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{6}} \:\mathrm{C}_{\mathrm{6}} ^{\mathrm{k}} \:\left(\mathrm{e}^{\mathrm{i}\theta} \right)^{\mathrm{k}} \:\left(\mathrm{e}^{−\mathrm{i}\theta} \right)^{\mathrm{6}−\mathrm{k}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{6}} \mathrm{C}_{\mathrm{6}} ^{\mathrm{k}} \:\mathrm{e}^{\mathrm{ik}\theta+\mathrm{ik}\theta−\mathrm{6i}\theta} \:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{6}} \:\mathrm{C}_{\mathrm{6}} ^{\mathrm{k}} \:\mathrm{e}^{\mathrm{i}\left(\mathrm{2k}−\mathrm{6}\right)\theta} \\ $$$$=..... \\ $$