Question and Answers Forum

IntegrationQuestion and Answers: Page 35

Question Number 178246 Answers: 1 Comments: 0

Question Number 177913 Answers: 0 Comments: 0

if I_n =∫xsin^n x dx and I_n = −((xsin^(n−1) x cosx )/n) +((sin^n x )/n^2 )+f(n)I_(n−2) then f(n) = ?

$if I_{n} = \int {xsin}^{n} x dx and$ $I_{n} = - \frac{{xsin}^{n - 1} x cosx}{n} + \frac{\sin^{n} x}{n^{2}} + f (n) I_{n - 2}$ $then f (n) = ?$

Question Number 177906 Answers: 1 Comments: 0

∫_0 ^1 ((sin x(cos^2 x−cos^2 (π/5))(cos^2 x−cos^2 ((2π)/5)))/(sin 5x)) dx =?

$\underset{0}{\int^{1}} \frac{\sin x (\cos^{2} x - \cos^{2} \frac{π}{5}) (\cos^{2} x - \cos^{2} \frac{2 π}{5})}{\sin 5 x} dx = ?$

Question Number 177820 Answers: 2 Comments: 0

∫_(π/2) ^π (dx/((sin x−2cos x)^2 )) =?

$\underset{π / 2}{\int^{π}} \frac{dx}{{(\sin x - 2 \cos x)}^{2}} = ?$

Question Number 177784 Answers: 1 Comments: 0

Question Number 177775 Answers: 1 Comments: 0

∫((sin2x)/( (√(cos^4 x+1))))dx

$\int \frac{\sin 2 x}{\sqrt{\cos^{4} x + 1}} d x$

Question Number 177604 Answers: 0 Comments: 0

Calculate 𝛗=∫_0 ^( 1) (( ln(x ))/(1 + x^( 2) )) dx =^? −G ∼ Solution ∼ 𝛗 = ∫_0 ^( 1) {Σ_(k=0) ^∞ (−1)^( k) x^( 2k) ln(x) }dx = Σ_(k=0) ^∞ (−1 )^( k) ∫_0 ^( 1) x^( 2k) ln(x) dx =^(i.b.p) Σ_(k=0) ^∞ (−1 )^( k) {[(x^( 1+2k) /(1+2k)) ln(x ) ]_0 ^1 −(1/(1+2k))∫_0 ^( 1) x^( 2k) dx } = Σ_(k=0) ^∞ (( (−1)^(1+k) )/(( 1 + 2k )^( 2) )) = − G ( Catalan constant ) ∴ 𝛗 = − G ■ m.n

$Calculate$ $ϕ = \int_{0}^{1} \frac{\ln (x)}{1 + x^{2}} d x \overset{?}{=} - G$ $\sim Solution \sim$ $ϕ = \int_{0}^{1} {\underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} x^{2 k} \ln (x)} d x$ $= \underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} \int_{0}^{1} x^{2 k} \ln (x) d x$ $\overset{i . b . p}{=} \underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} {{[\frac{x^{1 + 2 k}}{1 + 2 k} \ln (x)]}_{0}^{1} - \frac{1}{1 + 2 k} \int_{0}^{1} x^{2 k} d x}$ $= \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{1 + k}}{{(1 + 2 k)}^{2}} = - G (C a t a l a n c o n s t a n t)$ $∴ ϕ = - G ◼ m . n$

Question Number 177541 Answers: 0 Comments: 3

Question Number 178487 Answers: 3 Comments: 0

∫ (dx/(cot^3 x sin^7 x)) =?

$\int \frac{d x}{\cot^{3} x \sin^{7} x} = ?$

Question Number 177298 Answers: 1 Comments: 0

Prove that ∫_0 ^(π/2) ((sin^2 x)/((sin x+cos x)))dx=(1/( (√2)))log ((√2)+1)

$Prove that$ $\int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{(\sin x + \cos x)} dx = \frac{1}{\sqrt{2}} \log (\sqrt{2} + 1)$

Question Number 177296 Answers: 2 Comments: 1

Evaluate ∫_0 ^π (x/(a^2 cos^2 x+b^2 sin^2 x))dx

$Evaluate$ $\int_{0}^{π} \frac{x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} dx$

Question Number 177242 Answers: 1 Comments: 0

∫ ((3x^(16) +5x^(14) )/((x^5 +x^2 +1)^4 ))dx

$\int \frac{{3 x}^{16} + {5 x}^{14}}{{(x^{5} + x^{2} + 1)}^{4}} dx$

Question Number 177241 Answers: 1 Comments: 0

∫ ((sin x)/(sin (x−a)))dx

$\int \frac{\sin x}{\sin (x - a)} dx$

Question Number 177194 Answers: 1 Comments: 0

∫ (dx/((sin x)^((14)/9) (cos x)^(4/9) ))

$\int \frac{dx}{{(\sin x)}^{\frac{14}{9}} {(\cos x)}^{\frac{4}{9}}}$

Question Number 176991 Answers: 2 Comments: 0

Question Number 176964 Answers: 0 Comments: 0

Question Number 176679 Answers: 2 Comments: 0

Eeasy integral.... 𝛀 = ∫_(−∫_0 ^( ∞) e^( −x^( 2) ) dx) ^( ∫_0 ^( ∞) e^( −x^( 2) ) dx) sin^( 2) (t).ln^( 3) ( t + (√(1+t^( 2) )))dt −−−m.n−−−

$E e a s y i n t e g r a l . . . .$ $Ω = \int_{- \int_{0}^{\infty} e^{- x^{2}} d x}^{\int_{0}^{\infty} e^{- x^{2}} d x} {s i n}^{2} (t) . {l n}^{3} (t + \sqrt{1 + t^{2}}) d t$ $- - - m . n - - -$

Question Number 176637 Answers: 0 Comments: 0

Question Number 176594 Answers: 0 Comments: 1

𝛗=∫_0 ^( 1) (( ( tanh^( −1) (x))^2 )/((1+x )^( 2) )) dx = ? ≺ solution ≻ note : tanh^( −1) (x)=− (1/2) ln(((1−x)/(1+x))) 𝛗= (1/4)∫_0 ^( 1) (( ln^( 2) (((1−x)/(1+x)) ))/((1+x )^( 2) )) dx =^(((1−x)/(1+x)) = t) (1/8)∫_0 ^( 1) ln^( 2) (t )dt =(1/8_ ) { [t.ln^( 2) (t)]_0 ^( 1) −2∫_0 ^( 1) ln(t)dt} =− (1/4) ∫_0 ^( 1) ln(t)dt= (1/4) ◂ m.n ▶

$ϕ = \int_{0}^{1} \frac{{({t a n h}^{- 1} (x))}^{2}}{{(1 + x)}^{2}} d x = ?$ $≺ s o l u t i o n ≻$ $n o t e : {t a n h}^{- 1} (x) = - \frac{1}{2} l n (\frac{1 - x}{1 + x})$ $ϕ = \frac{1}{4} \int_{0}^{1} \frac{{l n}^{2} (\frac{1 - x}{1 + x})}{{(1 + x)}^{2}} d x$ $\overset{\frac{1 - x}{1 + x} = t}{=} \frac{1}{8} \int_{0}^{1} {l n}^{2} (t) d t$ $= \frac{1}{8_{}} {{[t . {l n}^{2} (t)]}_{0}^{1} - 2 \int_{0}^{1} l n (t) d t}$ $= - \frac{1}{4} \int_{0}^{1} l n (t) d t = \frac{1}{4} ◂ m . n ▸$

Question Number 176570 Answers: 2 Comments: 0

(1) ∫^(π/2) _(π/3) ((1+sinx)/(cosx)) dx=?

$(1) {\int_{\frac{π}{3}}}^{\frac{π}{2}} \frac{1 + s i n x}{c o s x} d x = ?$

Question Number 176566 Answers: 0 Comments: 0

−−−− calculate: Φ = Σ_(n=0) ^( ∞) (( 1)/((2n+1 ).e^( 4n+2) )) = ? where ” e ” is euler number. ≺ solution ≻ Φ = Σ_(n=0) ^∞ (1/e^( 4n+2) ) ∫_0 ^( 1) x^( 2n) dx = (1/e^( 2) ) ∫_0 ^( 1) Σ_(n=0) ^∞ ((( x^2 )/e^( 4) ) )^( n) dx = (1/e^( 2) ) ∫_0 ^( 1) (( 1)/(1− ((x/e^( 2) ) )^( 2) )) dx=(1/(2e^( 2) )) ∫_0 ^( 1) (1/(1−(x/e^( 2) ))) +(1/(1+(x/e^( 2) )))dx = (1/2) ln ( ((1+(1/e^( 2) ))/(1−(1/e^( 2) ))) ) = tanh^( −1) ((( 1)/e^( 2) ) ) ∴ Φ = coth^( −1) ( e^( 2) ) ■ m.n

$- - - -$ $c a l c u l a t e : Φ = \underset{n = 0}{\sum^{\infty}} \frac{1}{(2 n + 1) . e^{4 n + 2}} = ?$ $w h e r e^{″} e^{″} i s e u l e r n u m b e r .$ $≺ s o l u t i o n ≻$ $Φ = \underset{n = 0}{\sum^{\infty}} \frac{1}{e^{4 n + 2}} \int_{0}^{1} x^{2 n} d x$ $= \frac{1}{e^{2}} \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} {(\frac{x^{2}}{e^{4}})}^{n} d x$ $= \frac{1}{e^{2}} \int_{0}^{1} \frac{1}{1 - {(\frac{x}{e^{2}})}^{2}} d x = \frac{1}{2 e^{2}} \int_{0}^{1} \frac{1}{1 - \frac{x}{e^{2}}} + \frac{1}{1 + \frac{x}{e^{2}}} d x$ $= \frac{1}{2} l n (\frac{1 + \frac{1}{e^{2}}}{1 - \frac{1}{e^{2}}}) = {t a n h}^{- 1} (\frac{1}{e^{2}})$ $∴ Φ = {c o t h}^{- 1} (e^{2}) ◼ m . n$

Question Number 176542 Answers: 1 Comments: 0

Ω = ∫_0 ^( 1) (( x.tanh^( −1) (x))/((1+x)^( 2) ))dx= (1/(24)) (π^( 2) −6)

$Ω = \int_{0}^{1} \frac{x . {t a n h}^{- 1} (x)}{{(1 + x)}^{2}} d x = \frac{1}{24} (π^{2} - 6)$

Question Number 176421 Answers: 2 Comments: 0

Question Number 176334 Answers: 1 Comments: 0

Ψ = ∫_0 ^( 1) (( ln( 1+ x − x^( 2) ))/x)dx = ?

$Ψ = \int_{0}^{1} \frac{\ln (1 + x - x^{2})}{x} d x = ?$

Question Number 176289 Answers: 0 Comments: 1

∫ ((log (cosx + (√(cos2x))) )/(1−cos^2 x ))dx

$\int \frac{\log (\cos x + \sqrt{\cos 2 x})}{1 - \cos^{2} x} d x$

Question Number 176288 Answers: 0 Comments: 0

∫cos2xlog(1+tanx)dx

$\int \cos 2 x \log (1 + \tan x) d x$

Pg 30 Pg 31 Pg 32 Pg 33 Pg 34 Pg 35 Pg 36 Pg 37 Pg 38 Pg 39

Contact: info@tinkutara.com