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Question Number 53295 Answers: 1 Comments: 1

∫_0 ^(π/2) (1/(2+cos x)) dx=...

$\int_{0}^{\frac{π}{2}} \frac{1}{2 + \cos x} d x = . . .$

Question Number 53294 Answers: 1 Comments: 0

∫_(−1/2) ^(1/2) [(((x+1)/(x−1)))^2 +(((x−1)/(x+1)))^2 −2]^(1/2) dx=...

$\int_{- 1 / 2}^{1 / 2} {[{(\frac{x + 1}{x - 1})}^{2} + {(\frac{x - 1}{x + 1})}^{2} - 2]}^{1 / 2} d x = . . .$

Question Number 53293 Answers: 1 Comments: 1

∫_(−1/2) ^(1/2) ∣xcos ((πx)/2)∣ dx=...

$\int_{- 1 / 2}^{1 / 2} ∣ x \cos \frac{π x}{2} ∣ d x = . . .$

Question Number 53292 Answers: 1 Comments: 1

∫_0 ^1 e^x^2 dx=..

$\int_{0}^{1} e^{x^{2}} d x = . .$

Question Number 53285 Answers: 0 Comments: 0

let I_λ =∫_0 ^π ((xdx)/(cos^2 x +λ^2 sin^2 x)) with λ real 1) find the value of I_λ 2) calculate ∫_0 ^π ((xdx)/(a^2 cos^2 x +b^2 sin^2 x)) with a and b reals.

$l e t I_{λ} = \int_{0}^{π} \frac{x d x}{{c o s}^{2} x + λ^{2} {s i n}^{2} x} w i t h λ r e a l$ $1) f i n d t h e v a l u e o f I_{λ}$ $2) c a l c u l a t e \int_{0}^{π} \frac{x d x}{a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x} w i t h a a n d b r e a l s .$

Question Number 53284 Answers: 0 Comments: 2

find f(x)=∫_0 ^∞ ((arctan(xt))/(1+t^2 ))dt with x real .

$f i n d f (x) = \int_{0}^{\infty} \frac{a r c t a n (x t)}{1 + t^{2}} d t w i t h x r e a l .$

Question Number 53271 Answers: 0 Comments: 2

1)calculate∫_0 ^∞ e^(−xt^2 ) dt with x>0 2) find the value of ∫_0 ^∞ ((e^(−t^2 ) −e^(−2t^2 ) )/t^2 ) dt by using fubinni theorem .

$1) c a l c u l a t e \int_{0}^{\infty} e^{- {x t}^{2}} d t w i t h x > 0$ $2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{e^{- t^{2}} - e^{- 2 t^{2}}}{t^{2}} d t b y u s i n g$ $f u b i n n i t h e o r e m .$

Question Number 53270 Answers: 1 Comments: 1

1)calculate ∫_0 ^∞ e^(−at) dt with a>0 2)by using fubinni theorem find the value of ∫_0 ^∞ ((e^(−t) −e^(−xt) )/t)dt with x>0 .

$1) c a l c u l a t e \int_{0}^{\infty} e^{- a t} d t w i t h a > 0$ $2) b y u s i n g f u b i n n i t h e o r e m f i n d t h e v a l u e o f$ $\int_{0}^{\infty} \frac{e^{- t} - e^{- x t}}{t} d t w i t h x > 0 .$

Question Number 53261 Answers: 0 Comments: 0

1)find f(x)=∫_0 ^1 e^(−2t) ln(1−xt)dt with ∣x∣<1 2) calculate ∫_0 ^1 e^(−2t) ln(1−((t(√2))/2))dt.

$1) f i n d f (x) = \int_{0}^{1} e^{- 2 t} l n (1 - x t) d t w i t h ∣ x ∣< 1$ $2) c a l c u l a t e \int_{0}^{1} e^{- 2 t} l n (1 - \frac{t \sqrt{2}}{2}) d t .$

Question Number 53259 Answers: 1 Comments: 0

Question Number 53228 Answers: 0 Comments: 3

1) find f(a) =∫_0 ^1 (dx/((ax+1)(√(x^2 −x+1)))) with a>0 2) calculate f^′ (a) 3)find the value of ∫_0 ^1 ((xdx)/((ax+1)^2 (√(x^2 −x+1)))) 4) calculate ∫_0 ^1 (dx/((2x+1)(√(x^2 −x+1)))) and ∫_0 ^1 ((xdx)/((2x+1)^2 (√(x^2 −x+1))))

$1) f i n d f (a) = \int_{0}^{1} \frac{d x}{(a x + 1) \sqrt{x^{2} - x + 1}} w i t h a > 0$ $2) c a l c u l a t e f^{^{'}} (a)$ $3) f i n d t h e v a l u e o f \int_{0}^{1} \frac{x d x}{{(a x + 1)}^{2} \sqrt{x^{2} - x + 1}}$ $4) c a l c u l a t e \int_{0}^{1} \frac{d x}{(2 x + 1) \sqrt{x^{2} - x + 1}} a n d \int_{0}^{1} \frac{x d x}{{(2 x + 1)}^{2} \sqrt{x^{2} - x + 1}}$

Question Number 53212 Answers: 2 Comments: 21

Let f(x) = ((2x)/(x^2 + 4)) (a) Find ∫_(−b) ^b f(x) dx, for b > 0 (b) Determine ∫_(−∞) ^∞ f(x) dx is convergent or not

$Let f (x) = \frac{2 x}{x^{2} + 4}$ $(a) Find \underset{- b}{\int^{b}} f (x) d x, for b > 0$ $(b) Determine \underset{- \infty}{\int^{\infty}} f (x) d x is convergent or not$

Question Number 53119 Answers: 6 Comments: 3

Evaluate : 1) ∫(√((2−x)/(4+x))) dx 2) ∫ (√((x−2)/(x−4))) dx 3) ∫ (√((x−2)(x−4))) dx 4) ∫ (dx/(2sinx+3secx)) .

$E v a l u a t e :$ $1) \int \sqrt{\frac{2 - x}{4 + x}} d x$ $2) \int \sqrt{\frac{x - 2}{x - 4}} d x$ $3) \int \sqrt{(x - 2) (x - 4)} d x$ $4) \int \frac{d x}{2 \sin x + 3 \sec x} .$

Question Number 53118 Answers: 1 Comments: 0

If a<∫_0 ^(2π) (1/(10+3 cos x)) dx<b, then the ordered pair (a, b) is

$If a < \int_{0}^{2 π} \frac{1}{10 + 3 \cos x} d x < b, then the$ $ordered pair (a, b) is$

Question Number 53114 Answers: 0 Comments: 1

let A_n =∫_0 ^∞ ((x sin(nx))/((x^2 +n^2 )^2 ))dx with n integr natural not 0 1) find the value of A_n 2) study the convergence of Σ A_n

$l e t A_{n} = \int_{0}^{\infty} \frac{x s i n (n x)}{{(x^{2} + n^{2})}^{2}} d x w i t h n i n t e g r n a t u r a l n o t 0$ $1) f i n d t h e v a l u e o f A_{n}$ $2) s t u d y t h e c o n v e r g e n c e o f Σ A_{n}$

Question Number 53113 Answers: 0 Comments: 1

let I =∫_(−∞) ^(+∞) ((t+1)/((t^2 −t+1)^2 ))dt find value of I .

$l e t I = \int_{- \infty}^{+ \infty} \frac{t + 1}{{(t^{2} - t + 1)}^{2}} d t$ $f i n d v a l u e o f I .$

Question Number 53112 Answers: 1 Comments: 0

calculate ∫_0 ^π ((1+2sinx)/(3 +2cosx))dx let A =∫_0 ^π ((1+2sinx)/(3 +2cosx))dx changement tan((x/2))=t give A =∫_0 ^∞ ((1+((4t)/(1+t^2 )))/(3+2((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =2 ∫_0 ^∞ ((1+t^2 +4t)/((1+t^2 )^2 (((3+3t^2 +2−2t^2 )/(1+t^2 )))))dt =2 ∫_0 ^∞ ((t^2 +4t +1)/((1+t^2 )(5+t^2 )))dt let decompose F(t)=((t^2 +4t+1)/((t^2 +1)(t^2 +5))) F(t)=((at +b)/(t^2 +1)) +((ct +d)/(t^2 +5)) ⇒(at+b)(t^2 +5)+(ct+d)(t^2 +1) =t^2 +4t +1 ⇒ at^3 +5at +bt^2 +5b +ct^3 +ct +dt^2 +d =t^2 +4t +1 ⇒ (a+c)t^3 +(b+d)t^2 +(5a+c)t +5b +d =t^2 +4t +1 ⇒a+c=0 and b+d=1 and 5a+c =4 and 5b+d =1 ⇒c=−a ⇒a=1 ⇒c=−1 we have d=1−b ⇒5b +1−b =1 ⇒b=0 ⇒d=1 ⇒ F(t)=(t/(t^2 +1)) +((−t +1)/(t^2 +5)) ⇒ A =2 ∫_0 ^∞ F(t)dt =∫_0 ^∞ ((2t)/(t^2 +1))dt +∫_0 ^∞ ((−2t +2)/(t^2 +5))dt =[ln(((t^2 +1)/(t^2 +5)))]_0 ^(+∞) +2 ∫_0 ^∞ (dt/(t^2 +5)) =ln(5) + 2 ∫_0 ^∞ (dt/(t^2 +5)) but ∫_0 ^∞ (dt/(t^2 +5))dt =_(t =(√5)u ) ∫_0 ^∞ (((√5)du)/(5(1+u^2 ))) =(1/(√5)) [artanu]_0 ^(+∞) =(π/(2(√5))) ⇒ A =ln(5) +(π/(2(√5))) .