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Question Number 61535 Answers: 0 Comments: 0

calculate ∫_0 ^(π/2) ((ln(1+cosx))/(cosx)) dx

$c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{l n (1 + c o s x)}{c o s x} d x$

Question Number 61534 Answers: 0 Comments: 0

calculate f(a) =∫∫_W (x+ay)e^(−x) e^(−ay) dxdy with W_a ={(x,y)∈R^2 /x≥0 ,y≥0 , x+ay ≤1 } a>0

$c a l c u l a t e f (a) = \int \int_{W} (x + a y) e^{- x} e^{- a y} d x d y w i t h$ $W_{a} = {(x, y) \in R^{2} / x ⩾ 0, y ⩾ 0, x + a y ⩽ 1} a > 0$

Question Number 61533 Answers: 0 Comments: 2

∫∫_([0,1]^2 ) ((x−y)/((x^2 +3y^(2 ) +1)^2 )) dxdy

$\int \int_{{[0, 1]}^{2}} \frac{x - y}{{(x^{2} + 3 y^{2} + 1)}^{2}} d x d y$

Question Number 61532 Answers: 0 Comments: 0

prove that (((a+b)^n )/(a^n +b^n )) <2^(n−1) ∀ n>1 (n natural)

$p r o v e t h a t \frac{{(a + b)}^{n}}{a^{n} + b^{n}} < 2^{n - 1} \forall n > 1 (n n a t u r a l)$

Question Number 61530 Answers: 0 Comments: 5

let U_n =∫_0 ^∞ (x^(−2n) /(1+x^4 )) dx with n integr natural and n≥1 1) calculate U_n interms of n 2) find lim_(n→+∞) n^2 U_n 3) study the serie Σ U_n

$l e t U_{n} = \int_{0}^{\infty} \frac{x^{- 2 n}}{1 + x^{4}} d x w i t h n i n t e g r n a t u r a l a n d n ⩾ 1$ $1) c a l c u l a t e U_{n} i n t e r m s o f n$ $2) f i n d {l i m}_{n \to + \infty} n^{2} U_{n}$ $3) s t u d y t h e s e r i e Σ U_{n}$

Question Number 61529 Answers: 0 Comments: 0

find ∫_0 ^∞ x^2 e^(−zx^2 ) dx with z from C

$f i n d \int_{0}^{\infty} x^{2} e^{- {z x}^{2}} d x w i t h z f r o m C$

Question Number 61528 Answers: 0 Comments: 4

find ∫_0 ^∞ cos(zx^2 )dx with z ∈ C .

$f i n d \int_{0}^{\infty} c o s ({z x}^{2}) d x w i t h z \in C .$

Question Number 61526 Answers: 0 Comments: 0

Solve for n: D/A×{1−((P×((((1+i)^n ×i)/((1+i)^n −1))))/((P×((((1+i)^r ×i)/((1+i)^r −1))))−(R/i)×[((1/n)+i)×((((1+i)^r ×i)/((1+i)^r −1)))−((1/n)+i)×((((1+i)^n ×i)/((1+i)^n −1)))]))}−1=0

$S o l v e f o r n : D / A \times {1 - \frac{P \times (\frac{{(1 + i)}^{n} \times i}{{(1 + i)}^{n} - 1})}{(P \times (\frac{{(1 + i)}^{r} \times i}{{(1 + i)}^{r} - 1})) - \frac{R}{i} \times [(\frac{1}{n} + i) \times (\frac{{(1 + i)}^{r} \times i}{{(1 + i)}^{r} - 1}) - (\frac{1}{n} + i) \times (\frac{{(1 + i)}^{n} \times i}{{(1 + i)}^{n} - 1})]}} - 1 = 0$

Question Number 61522 Answers: 1 Comments: 0

I=∫((sin x.e^(cos x) −(sin x+cos x)e^((sin x+cos x)) )/(e^(2sin x) −2e^(sin x) +1))dx

$I = \int \frac{\sin x . e^{\cos x} - (\sin x + \cos x) e^{(\sin x + \cos x)}}{e^{2 \sin x} - 2 e^{\sin x} + 1} d x$

Question Number 61521 Answers: 1 Comments: 0

Question Number 61520 Answers: 0 Comments: 0

Question Number 61495 Answers: 1 Comments: 7

Question Number 61511 Answers: 1 Comments: 0

cos^(−1) ((2x^2 −1)/(2x^2 )) + cos^(−1) ((x^2 −2)/x^2 )=120°=((2π)/3) Find x

${c o s}^{- 1} \frac{2 x^{2} - 1}{2 x^{2}} + {c o s}^{- 1} \frac{x^{2} - 2}{x^{2}} = 120 ° = \frac{2 π}{3}$ $Find x$

Question Number 61510 Answers: 0 Comments: 0

S_1 =Σ_(k=1) ^n (√((16n−16k)(16n+16k))) S_2 =Σ_(k=1) ^n (√((16k−16)(16k+16))) lim_(n→∞) ((S_1 +S_2 )/n^2 )=?

$S_{1} = \underset{k = 1}{\sum^{n}} \sqrt{(16 n - 16 k) (16 n + 16 k)}$ $S_{2} = \underset{k = 1}{\sum^{n}} \sqrt{(16 k - 16) (16 k + 16)}$ $\lim_{n \to \infty} \frac{S_{1} + S_{2}}{n^{2}} = ?$

Question Number 61490 Answers: 0 Comments: 10

(√(a−(√(a+x))))+(√(a+(√(a−x))))=2x this is the solution Sir Aifour and me found trivial solution a=x=0 a, x ∈R x=((√2)/8)(r+(√(r^2 +4)))(√(2(4a−1)−r^2 −r(√(r^2 +4)))) with r=−2(√((4a−3)/3))sin ((1/3)arcsin ((3(√3))/((4a−3)^(3/2) ))) no solution for a<a_0 with a_0 ≈1.509830340886

$\sqrt{a - \sqrt{a + x}} + \sqrt{a + \sqrt{a - x}} = 2 x$ $this is the solution Sir Aifour and me found$ $trivial solution a = x = 0$ $a, x \in R$ $x = \frac{\sqrt{2}}{8} (r + \sqrt{r^{2} + 4}) \sqrt{2 (4 a - 1) - r^{2} - r \sqrt{r^{2} + 4}}$ $with$ $r = - 2 \sqrt{\frac{4 a - 3}{3}} \sin (\frac{1}{3} \arcsin \frac{3 \sqrt{3}}{{(4 a - 3)}^{\frac{3}{2}}})$ $no solution for a < a_{0} with a_{0} \approx 1.509830340886$