Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 36762 by abdo.msup.com last updated on 05/Jun/18

find A_n = ∫_0 ^(π/4) (cosx +sinx)^n dx.

$${find}\:{A}_{{n}} \:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left({cosx}\:+{sinx}\right)^{{n}} \:{dx}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18

Commented by abdo.msup.com last updated on 05/Jun/18

A_n = ∫_0 ^(π/4) {(√2)cos(x−(π/4))}^n dx =2^(n/2) ∫_0 ^(π/4) cos^n (x−(π/4))dx changemrnt x−(π/4)=−t A_n =2^(n/2) ∫_(π/4) ^0 cos^n (t)(−dt) =2^(n/2) ∫_0 ^(π/4) cos^n t dt =_(t=2x) 2((√2))^n ∫_0 ^(π/2) cos^n (2x)dx =2 ((√2))^n ∫_0 ^(π/2) (2cos^2 x−1)^n dx =2((√2)) ∫_0 ^(π/2) Σ_(k=0) ^n C_n ^k 2^k cos^(2k) (x)(−1)^(n−k) dx =2((√2))^n Σ_(k=0) ^n C_n ^k 2^k (−1)^(n−k) ∫_0 ^(π/2) cos^(2k) x dx =2((√2))^n Σ_(k=0) ^n (−1)^(n−k) 2^k C_n ^k W_k with W_k = ∫_0 ^(π/2) cos^(2k) xdx the value of W_k is known(wsllis integral) .

$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left\{\sqrt{\mathrm{2}}{cos}\left({x}−\frac{\pi}{\mathrm{4}}\right)\right\}^{{n}} {dx} \\ $$$$=\mathrm{2}^{\frac{{n}}{\mathrm{2}}} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{{n}} \left({x}−\frac{\pi}{\mathrm{4}}\right){dx}\:{changemrnt} \\ $$$${x}−\frac{\pi}{\mathrm{4}}=−{t} \\ $$$${A}_{{n}} \:=\mathrm{2}^{\frac{{n}}{\mathrm{2}}} \:\int_{\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} \:{cos}^{{n}} \left({t}\right)\left(−{dt}\right) \\ $$$$=\mathrm{2}^{\frac{{n}}{\mathrm{2}}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{{n}} {t}\:{dt} \\ $$$$=_{{t}=\mathrm{2}{x}} \:\mathrm{2}\left(\sqrt{\mathrm{2}}\right)^{{n}} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{cos}^{{n}} \left(\mathrm{2}{x}\right){dx} \\ $$$$=\mathrm{2}\:\left(\sqrt{\mathrm{2}}\right)^{{n}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\right)^{{n}} {dx} \\ $$$$=\mathrm{2}\left(\sqrt{\mathrm{2}}\right)\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\mathrm{2}^{{k}} \:{cos}^{\mathrm{2}{k}} \left({x}\right)\left(−\mathrm{1}\right)^{{n}−{k}} {dx} \\ $$$$=\mathrm{2}\left(\sqrt{\mathrm{2}}\right)^{{n}} \:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \mathrm{2}^{{k}} \left(−\mathrm{1}\right)^{{n}−{k}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{k}} {x}\:{dx} \\ $$$$=\mathrm{2}\left(\sqrt{\mathrm{2}}\right)^{{n}} \:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{n}−{k}} \:\mathrm{2}^{{k}} \:{C}_{{n}} ^{{k}} \:\:{W}_{{k}} \\ $$$${with}\:{W}_{{k}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{k}} {xdx}\:{the}\:{value}\:{of} \\ $$$${W}_{{k}} \:{is}\:{known}\left({wsllis}\:{integral}\right)\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18

A_n =∫_0 ^(Π/4) {(√2) ((1/(√2))cosx+(1/((√2) ))sinx)}^n dx A_n =∫_0 ^(Π/4) 2^(n/2) ×sin^n ((Π/4)+x)dx t=(Π/4)+x dt=dx A_n =2^(n/2) ∫_(Π/4) ^(Π/2) sin^n tdt I_n =∫sin^n tdt use reduction formula contd

$${A}_{{n}} =\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} \left\{\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{cosx}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}{sinx}\right)\right\}^{{n}} {dx} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} \mathrm{2}^{\frac{{n}}{\mathrm{2}}} ×{sin}^{{n}} \left(\frac{\Pi}{\mathrm{4}}+{x}\right){dx} \\ $$$${t}=\frac{\Pi}{\mathrm{4}}+{x} \\ $$$${dt}={dx} \\ $$$${A}_{{n}} =\mathrm{2}^{\frac{{n}}{\mathrm{2}}} \int_{\frac{\Pi}{\mathrm{4}}} ^{\frac{\Pi}{\mathrm{2}}} {sin}^{{n}} {tdt} \\ $$$${I}_{{n}} =\int{sin}^{{n}} {tdt}\:\:\:{use}\:{reduction}\:{formula} \\ $$$${contd} \\ $$$$\: \\ $$$$ \\ $$$$\:\: \\ $$

Commented by NECx last updated on 05/Jun/18

wow.... i had never thouvht of this.

$${wow}....\:{i}\:{had}\:{never}\:{thouvht}\:{of} \\ $$$${this}. \\ $$

Commented by MJS last updated on 05/Jun/18

the method is ok but something seems not clear to me. it should be: for n=2k A_0 =(π/4); A_(k+1) =((2k−1)/k)A_k +(1/(2k)) for n=2k+1 A_1 =1; A_(k+1) =((4k)/(2k+1))A_k +(1/(2k+1)) ⇒ A_0 =(π/4); A_1 =1; A_(n+2) =2((n−1)/n)A_n +(1/n) please correct me if I′m wrong

$$\mathrm{the}\:\mathrm{method}\:\mathrm{is}\:\mathrm{ok}\:\mathrm{but}\:\mathrm{something}\:\mathrm{seems}\:\mathrm{not} \\ $$$$\mathrm{clear}\:\mathrm{to}\:\mathrm{me}.\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}: \\ $$$$ \\ $$$$\mathrm{for}\:{n}=\mathrm{2}{k} \\ $$$${A}_{\mathrm{0}} =\frac{\pi}{\mathrm{4}};\:{A}_{{k}+\mathrm{1}} =\frac{\mathrm{2}{k}−\mathrm{1}}{{k}}{A}_{{k}} +\frac{\mathrm{1}}{\mathrm{2}{k}} \\ $$$$\mathrm{for}\:{n}=\mathrm{2}{k}+\mathrm{1} \\ $$$${A}_{\mathrm{1}} =\mathrm{1};\:{A}_{{k}+\mathrm{1}} =\frac{\mathrm{4}{k}}{\mathrm{2}{k}+\mathrm{1}}{A}_{{k}} +\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}} \\ $$$$ \\ $$$$\Rightarrow\:{A}_{\mathrm{0}} =\frac{\pi}{\mathrm{4}};\:{A}_{\mathrm{1}} =\mathrm{1};\:{A}_{{n}+\mathrm{2}} =\mathrm{2}\frac{{n}−\mathrm{1}}{{n}}{A}_{{n}} +\frac{\mathrm{1}}{{n}} \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{correct}\:\mathrm{me}\:\mathrm{if}\:\mathrm{I}'\mathrm{m}\:\mathrm{wrong} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18

pls show some steps to understand...you also right..

$${pls}\:{show}\:{some}\:{steps}\:{to}\:{understand}...{you}\:{also} \\ $$$${right}.. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com