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Question Number 160618 Answers: 1 Comments: 0

Question Number 160615 Answers: 0 Comments: 1

Question Number 160609 Answers: 4 Comments: 0

lim_(x→π) (((tan x)/(1+cos x)))=?

$\lim_{x \to π} (\frac{\tan x}{1 + \cos x}) = ?$

Question Number 160608 Answers: 0 Comments: 6

Question Number 160606 Answers: 0 Comments: 1

Evaluate (d^2 y/dx^2 ) when x=ln 2 and y=2.

$Evaluate \frac{d^{2} y}{{d x}^{2}} when x = \ln 2 and y = 2 .$

Question Number 160605 Answers: 0 Comments: 1

Given that y=(3 sin x−4 cos x+6)^2 , 0≤x≤2π. Find the smallest value of y.

$Given that y = {(3 \sin x - 4 \cos x + 6)}^{2}, 0 ⩽ x ⩽ 2 π .$ $Find the smallest value of y .$

Question Number 160604 Answers: 0 Comments: 0

S_n = ((12)/((4^2 −3^2 )(4^2 −3^2 )))+((12^2 )/((4^2 −3^2 )(4^3 −3^3 )))+((12^3 )/((4^2 −3^2 )(4^4 −3^4 )))+...+((12^n )/((4^2 −3^2 )(4^(n+1) −3^(n+1) ))) lim_(n→∞) S_n = ?

$S_{n} = \frac{12}{(4^{2} - 3^{2}) (4^{2} - 3^{2})} + \frac{12^{2}}{(4^{2} - 3^{2}) (4^{3} - 3^{3})} + \frac{12^{3}}{(4^{2} - 3^{2}) (4^{4} - 3^{4})} + . . . + \frac{12^{n}}{(4^{2} - 3^{2}) (4^{n + 1} - 3^{n + 1})}$ $\lim_{n \to \infty} S_{n} = ?$

Question Number 160603 Answers: 1 Comments: 0

(x+(√(x^2 +1)))(y+(√(y^2 +1)))=2021 ∀x,y∈R^+ . min (x+y)=?

$(x + \sqrt{x^{2} + 1}) (y + \sqrt{y^{2} + 1}) = 2021$ $\forall x, y \in R^{+} . \min (x + y) = ?$

Question Number 160597 Answers: 1 Comments: 0

∫_0 ^( (π/2)) (dx/(2−cos x)) =?

$\int_{0}^{\frac{π}{2}} \frac{dx}{2 - \cos x} = ?$

Question Number 160596 Answers: 1 Comments: 0

Question Number 160594 Answers: 1 Comments: 0

∫ (dx/(1−tan^2 (x))) =?

$\int \frac{dx}{1 - \tan^{2} (x)} = ?$

Question Number 160626 Answers: 0 Comments: 1

∫ ((lnz)/(z+lnz)) dz help me sir

$\int \frac{l n z}{z + l n z} d z$ $h e l p m e s i r$

Question Number 160591 Answers: 0 Comments: 0

Question Number 160590 Answers: 0 Comments: 0

Ω:=∫_0 ^( (1/2)) (( arcsinh(x))/x) dx =^? (π^2 /(20))

$Ω := \int_{0}^{\frac{1}{2}} \frac{a r c s i n h (x)}{x} d x \overset{?}{=} \frac{π^{2}}{20}$

Question Number 160589 Answers: 0 Comments: 0

Question Number 160577 Answers: 1 Comments: 0

Ω=∫_0 ^( 1) tan^( −1) (x).ln(x) = ? −−−−solution−−−− f(a)=∫_0 ^( 1) tan^( −1) ( x) .x^( a) dx = Σ_(n=1) ^∞ (( (−1)^( n−1) )/(2n−1)) ∫_0 ^( 1) x^( 2n+a−1) dx = Σ_(n=0) ^∞ (((−1)^( n−1) )/(2n−1)) ((1/(2n+ a)) ) Ω= f ′ (a )∣_(a=0) = Σ_(n=1) ^∞ (((−1)^n )/((2n−1)( 2n+ a)^( 2) )) Ω= f ′ (0 )=(1/4)Σ(( (−1 )^(n−1) (2n−1−2n ) )/(( 2n−1)n^( 2) )) =(1/4) Σ_(n=1) ^∞ (((−1)^(n−1) )/n^( 2) ) − Σ_(n=1) ^∞ (((−1)^( n−1) )/((2n−1)(2n))) = (π^( 2) /(48)) −{ Σ_(n=1) ^∞ {(((−1)^(n−1) )/(2n−1)) −(((−1)^( n−1) )/(2n))}} ∴ Ω = (π^( 2) /(48)) − (π/4) +(1/2) ln(2)

$Ω = \int_{0}^{1} {t a n}^{- 1} (x) . l n (x) = ?$ $- - - - s o l u t i o n - - - -$ $f (a) = \int_{0}^{1} {t a n}^{- 1} (x) . x^{a} d x$ $= \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{2 n - 1} \int_{0}^{1} x^{2 n + a - 1} d x$ $= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{2 n - 1} (\frac{1}{2 n + a})$ $Ω = f^{'} (a) ∣_{a = 0} = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n - 1) {(2 n + a)}^{2}}$ $Ω = f^{'} (0) = \frac{1}{4} Σ \frac{{(- 1)}^{n - 1} (2 n - 1 - 2 n)}{(2 n - 1) n^{2}}$ $= \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n^{2}} - \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{(2 n - 1) (2 n)}$ $= \frac{π^{2}}{48} - {\underset{n = 1}{\sum^{\infty}} {\frac{{(- 1)}^{n - 1}}{2 n - 1} - \frac{{(- 1)}^{n - 1}}{2 n}}}$ $∴ Ω = \frac{π^{2}}{48} - \frac{π}{4} + \frac{1}{2} l n (2)$