∫_0 ^π (√((1+cos2x)/2)) dx ∫


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Question Number 119681 by TANMAY PANACEA last updated on 26/Oct/20

$\int_{0}^{π} \sqrt{\frac{1 + c o s 2 x}{2}} d x$ $\int_{0}^{\infty} [{n e}^{- x}] d x$
	Answered by bemath last updated on 26/Oct/20
	$(√(((1+cos 2x)/2) )) = (√((1+2cos^2 x−1)/2)) = ∣ cos x ∣ → { ((cos x for 0≤x≤(π/2))),((−cos x for (π/2)≤x≤π)) :} I = ∫_0 ^(π/2) ∣cos x∣ dx + ∫_(π/2) ^π ∣cos x∣ dx I = sin x ]_0 ^(π/2) −(sin x)]_(π/2) ^π I=1−(0−1)=2$
	$\sqrt{\frac{1 + \cos 2 x}{2}} = \sqrt{\frac{1 + 2 \cos^{2} x - 1}{2}}$ $= ∣ \cos x ∣ \to {\begin{cases} \cos x f o r 0 ⩽ x ⩽ \frac{π}{2} \\ - \cos x f o r \frac{π}{2} ⩽ x ⩽ π \end{cases}$ $I = \underset{0}{\int^{\frac{π}{2}}} ∣ \cos x ∣ d x + \underset{\frac{π}{2}}{\int^{π}} ∣ \cos x ∣ d x$ ${I {= \sin x]}_{0}^{\frac{π}{2}} - (\sin x)]}_{\frac{π}{2}}^{π}$ $I = 1 - (0 - 1) = 2$
	Commented by TANMAY PANACEA last updated on 26/Oct/20

	$e x c e l l e n t s i r$
	Answered by mindispower last updated on 26/Oct/20
	$∫_0 ^∞ [ne^(−x) ]dx=u_n t=ne^(−x) ⇒x=−ln((t/n))⇒dx=−(1/t)dt =∫_0 ^n (([t])/t)dt =Σ_(k=0) ^(n−1) ∫_k ^(k+1) (([t]dt)/t)=Σ_(k≤n−1) ∫_k ^(k+1) (k/t)dt =Σ_(k≤n−1) kln(1+(1/k))=Σ_(k≤n−1) (kln(k+1)−kln(k)) =Σ_(k≤n−1) ((k+1)ln(k+1)−kln(k)−ln(k+1))=u_n n=0,u_0 =0 n≥1 u_n =nln(n)−ln[Π_(k≤n−1) (k+1)]=nln(n)−ln(n!) u_n = { ((0 ,n=0)),((nln(n)−ln(n!) ,n≥1)) :}$
	$\int_{0}^{\infty} [{n e}^{- x}] d x = u_{n}$ $t = {n e}^{- x} \Rightarrow x = - l n (\frac{t}{n}) \Rightarrow d x = - \frac{1}{t} d t$ $= \int_{0}^{n} \frac{[t]}{t} d t$ $= \underset{k = 0}{\sum^{n - 1}} \int_{k}^{k + 1} \frac{[t] d t}{t} = \sum_{k ⩽ n - 1} \int_{k}^{k + 1} \frac{k}{t} d t$ $= \sum_{k ⩽ n - 1} k l n (1 + \frac{1}{k}) = \sum_{k ⩽ n - 1} (k l n (k + 1) - k l n (k))$ $= \sum_{k ⩽ n - 1} ((k + 1) l n (k + 1) - k l n (k) - l n (k + 1)) = u_{n}$ $n = 0, u_{0} = 0$ $n ⩾ 1$ $u_{n} = n l n (n) - l n [\prod_{k ⩽ n - 1} (k + 1)] = n l n (n) - l n (n!)$ $u_{n} = {\begin{cases} 0, n = 0 \\ n l n (n) - l n (n!), n ⩾ 1 \end{cases}$
	Commented by TANMAY PANACEA last updated on 26/Oct/20

	$t h a n k y o u s i r$
	Commented by mindispower last updated on 26/Oct/20

	$w i t h e p l e a s u r h a v e a n i c e d a y$
	Answered by mathmax by abdo last updated on 26/Oct/20

	$A = \int_{0}^{π} \sqrt{\frac{1 + \cos (2 x)}{2}} dx \Rightarrow A = \int_{0}^{π} ∣ cosx ∣ dx$ $= \int_{0}^{\frac{π}{2}} cosx dx - \int_{\frac{π}{2}}^{π} cosx dx = {[sinx]}_{0}^{\frac{π}{2}} - {[sinx]}_{\frac{π}{2}}^{π}$ $= 1 - (- 1) = 2 \Rightarrow A = 2$
	Answered by mathmax by abdo last updated on 26/Oct/20
	$I_n =∫_0 ^∞ [ne^(−x) ]dx we do the changement ne^(−x) =t ⇒e^(−x) =(t/n) ⇒−x=ln((t/n)) ⇒x =−ln((t/n))=ln(n)−ln(t) (we suppose n≠0) ⇒I_n =∫_n ^0 [t](−(dt/t)) =∫_0 ^n (([t])/t)dt =Σ_(k=0) ^(n−1) ∫_k ^(k+1) (k/t)dt =Σ_(k=0) ^(n−1) k{ln(k+1)−ln(k)} =Σ_(k=1) ^(n−1) kln(((k+1)/k)) =Σ_(k=1) ^(n−1) k ln(1+(1/k)) =ln(2)+2ln(1+(1/2))+3ln(1+(1/3))+...+(n−1)ln(1+(1/(n−1))) ifn=0 we get I_n =0$
	$I_{n} = \int_{0}^{\infty} [{ne}^{- x}] dx we do the changement {ne}^{- x} = t \Rightarrow e^{- x} = \frac{t}{n}$ $\Rightarrow - x = \ln (\frac{t}{n}) \Rightarrow x = - \ln (\frac{t}{n}) = \ln (n) - \ln (t) (we suppose n \neq 0)$ $\Rightarrow I_{n} = \int_{n}^{0} [t] (- \frac{dt}{t}) = \int_{0}^{n} \frac{[t]}{t} dt = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} \frac{k}{t} dt$ $= \sum_{k = 0}^{n - 1} k {\ln (k + 1) - \ln (k)}$ $= \sum_{k = 1}^{n - 1} kln (\frac{k + 1}{k}) = \sum_{k = 1}^{n - 1} k \ln (1 + \frac{1}{k})$ $= \ln (2) + 2 \ln (1 + \frac{1}{2}) + 3 \ln (1 + \frac{1}{3}) + . . . + (n - 1) \ln (1 + \frac{1}{n - 1})$ $ifn = 0 we get I_{n} = 0$
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