Question Number 30521 by abdo imad last updated on 22/Feb/18
$$\left.\mathrm{1}\right)\:{find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{{n}} \:{cos}\left({narctanx}\right){dx} \\ $$$$\left.\mathrm{2}\right){find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \:}\right)^{\mathrm{3}} \:{cos}\left(\mathrm{3}\:{arctanx}\right){dx}\:. \\ $$
Commented by abdo imad last updated on 24/Feb/18
![1)let consider the polynomial t(x)=(1+ix)^n +(1−ix)^n t(x)= 2Re(1+ix)^n = 2((√(1+x^2 )) )^n cos(narctanx)⇒ ∫_0 ^1 ((√(1+x^2 )) )^n cos(narctanx)dx=(1/2)∫_0 ^1 t(x)dx from another side t(x)= Σ_(k=0) ^n C_n ^k (ix)^k +Σ_(k=0) ^n C_n ^k (−ix)^k = Σ_(k=0) ^n C_n ^k ((i)^k +(−i)^k )x^k =2Σ_(p=0) ^([(n/2)]) C_n ^(2p) (−1)^p x^(2p) ⇒ (1/2)∫_0 ^1 t(x)dx= Σ_(p=0) ^([(n/2)]) C_n ^(2p) (−1)^(p ) ∫_0 ^1 x^(2p) dx = Σ_(p=0) ^([(n/2)]) (−1)^p (C_n ^(2p) /(2p+1)) . 2)forn=3 we find ∫_0 ^1 ((√(1+x^2 ))) )^3 cos(3arctanx)dx=Σ_(p=0) ^1 (−1)^p (C_3 ^(2p) /(2p+1)) =C_3 ^0 −(C_3 ^2 /3)=1−(1/3) =(2/3) .](https://www.tinkutara.com/question/Q30701.png)
$$\left.\mathrm{1}\right){let}\:{consider}\:{the}\:{polynomial}\:{t}\left({x}\right)=\left(\mathrm{1}+{ix}\right)^{{n}} \:+\left(\mathrm{1}−{ix}\right)^{{n}} \\ $$$${t}\left({x}\right)=\:\mathrm{2}{Re}\left(\mathrm{1}+{ix}\right)^{{n}} =\:\mathrm{2}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)^{{n}} {cos}\left({narctanx}\right)\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)^{{n}} {cos}\left({narctanx}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}\left({x}\right){dx}\:{from}\:{another} \\ $$$${side}\:{t}\left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({ix}\right)^{{k}} \:+\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(−{ix}\right)^{{k}} \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(\left({i}\right)^{{k}} \:+\left(−{i}\right)^{{k}} \right){x}^{{k}} \:\:=\mathrm{2}\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}} \left(−\mathrm{1}\right)^{{p}} \:{x}^{\mathrm{2}{p}} \:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}\left({x}\right){dx}=\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}} \:\left(−\mathrm{1}\right)^{{p}\:} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}{p}} {dx} \\ $$$$=\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\left(−\mathrm{1}\right)^{{p}} \:\:\frac{{C}_{{n}} ^{\mathrm{2}{p}} }{\mathrm{2}{p}+\mathrm{1}}\:. \\ $$$$\left.\mathrm{2}\right){forn}=\mathrm{3}\:{we}\:{find}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \:\right)}\:\right)^{\mathrm{3}} \:{cos}\left(\mathrm{3}{arctanx}\right){dx}=\sum_{{p}=\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{p}} \:\:\frac{{C}_{\mathrm{3}} ^{\mathrm{2}{p}} }{\mathrm{2}{p}+\mathrm{1}} \\ $$$$={C}_{\mathrm{3}} ^{\mathrm{0}} \:−\frac{{C}_{\mathrm{3}} ^{\mathrm{2}} }{\mathrm{3}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\:=\frac{\mathrm{2}}{\mathrm{3}}\:. \\ $$
Commented by abdo imad last updated on 24/Feb/18
$${C}_{\mathrm{3}} ^{\mathrm{0}} \:−\frac{{C}_{\mathrm{3}} ^{\mathrm{2}} }{\mathrm{3}}=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{3}}=\mathrm{0} \\ $$