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


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

$f i n d A_{n} = \int_{0}^{\frac{π}{4}} {(c o s x + s i n x)}^{n} d x .$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18

Commented by abdo.msup.com last updated on 05/Jun/18
$A_n = ∫_0 ^(π/4) {(√2)cos(x−(π/4))}^n dx =2^(n/2) ∫_0 ^(π/4) cos^n (x−(π/4))dx changemrnt x−(π/4)=−t A_n =2^(n/2) ∫_(π/4) ^0 cos^n (t)(−dt) =2^(n/2) ∫_0 ^(π/4) cos^n t dt =_(t=2x) 2((√2))^n ∫_0 ^(π/2) cos^n (2x)dx =2 ((√2))^n ∫_0 ^(π/2) (2cos^2 x−1)^n dx =2((√2)) ∫_0 ^(π/2) Σ_(k=0) ^n C_n ^k 2^k cos^(2k) (x)(−1)^(n−k) dx =2((√2))^n Σ_(k=0) ^n C_n ^k 2^k (−1)^(n−k) ∫_0 ^(π/2) cos^(2k) x dx =2((√2))^n Σ_(k=0) ^n (−1)^(n−k) 2^k C_n ^k W_k with W_k = ∫_0 ^(π/2) cos^(2k) xdx the value of W_k is known(wsllis integral) .$
$A_{n} = \int_{0}^{\frac{π}{4}} {\sqrt{2} c o s (x - \frac{π}{4})}^{n} d x$ $= 2^{\frac{n}{2}} \int_{0}^{\frac{π}{4}} {c o s}^{n} (x - \frac{π}{4}) d x c h a n g e m r n t$ $x - \frac{π}{4} = - t$ $A_{n} = 2^{\frac{n}{2}} \int_{\frac{π}{4}}^{0} {c o s}^{n} (t) (- d t)$ $= 2^{\frac{n}{2}} \int_{0}^{\frac{π}{4}} {c o s}^{n} t d t$ $=_{t = 2 x} 2 {(\sqrt{2})}^{n} \int_{0}^{\frac{π}{2}} {c o s}^{n} (2 x) d x$ $= 2 {(\sqrt{2})}^{n} \int_{0}^{\frac{π}{2}} {(2 {c o s}^{2} x - 1)}^{n} d x$ $= 2 (\sqrt{2}) \int_{0}^{\frac{π}{2}} \sum_{k = 0}^{n} C_{n}^{k} 2^{k} {c o s}^{2 k} (x) {(- 1)}^{n - k} d x$ $= 2 {(\sqrt{2})}^{n} \sum_{k = 0}^{n} C_{n}^{k} 2^{k} {(- 1)}^{n - k} \int_{0}^{\frac{π}{2}} {c o s}^{2 k} x d x$ $= 2 {(\sqrt{2})}^{n} \sum_{k = 0}^{n} {(- 1)}^{n - k} 2^{k} C_{n}^{k} W_{k}$ $w i t h W_{k} = \int_{0}^{\frac{π}{2}} {c o s}^{2 k} x d x t h e v a l u e o f$ $W_{k} i s k n o w n (w s l l i s i n t e g r a l) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18
	$A_n =∫_0 ^(Π/4) {(√2) ((1/(√2))cosx+(1/((√2) ))sinx)}^n dx A_n =∫_0 ^(Π/4) 2^(n/2) ×sin^n ((Π/4)+x)dx t=(Π/4)+x dt=dx A_n =2^(n/2) ∫_(Π/4) ^(Π/2) sin^n tdt I_n =∫sin^n tdt use reduction formula contd$
	$A_{n} = \int_{0}^{\frac{Π}{4}} {\sqrt{2} (\frac{1}{\sqrt{2}} c o s x + \frac{1}{\sqrt{2}} s i n x)}^{n} d x$ $A_{n} = \int_{0}^{\frac{Π}{4}} 2^{\frac{n}{2}} \times {s i n}^{n} (\frac{Π}{4} + x) d x$ $t = \frac{Π}{4} + x$ $d t = d x$ $A_{n} = 2^{\frac{n}{2}} \int_{\frac{Π}{4}}^{\frac{Π}{2}} {s i n}^{n} t d t$ $I_{n} = \int {s i n}^{n} t d t u s e r e d u c t i o n f o r m u l a$ $c o n t d$
	Commented by NECx last updated on 05/Jun/18

	$w o w . . . . i h a d n e v e r t h o u v h t o f$ $t h i s .$
	Commented by MJS last updated on 05/Jun/18

	$the method is ok but something seems not$ $clear to me . it should be :$ $for n = 2 k$ $A_{0} = \frac{π}{4}; A_{k + 1} = \frac{2 k - 1}{k} A_{k} + \frac{1}{2 k}$ $for n = 2 k + 1$ $A_{1} = 1; A_{k + 1} = \frac{4 k}{2 k + 1} A_{k} + \frac{1}{2 k + 1}$ $\Rightarrow A_{0} = \frac{π}{4}; A_{1} = 1; A_{n + 2} = 2 \frac{n - 1}{n} A_{n} + \frac{1}{n}$ $please correct me if I^{'} m wrong$
	Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18

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

	$p l s s h o w s o m e s t e p s t o u n d e r s t a n d . . . y o u a l s o$ $r i g h t . .$
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