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IntegrationQuestion and Answers: Page 112

Question Number 128057 Answers: 3 Comments: 0

∫_0 ^( 1) ∫_0 ^( 1) (1/(1−xy^2 )) dx dy =?

$\int_{0}^{1} \int_{0}^{1} \frac{1}{1 - {xy}^{2}} dx dy = ?$

Question Number 128052 Answers: 3 Comments: 0

I = ∫ arctan (((x−2)/(x+2))) dx

$I = \int \arctan (\frac{x - 2}{x + 2}) dx$

Question Number 128026 Answers: 0 Comments: 0

∫_(π/2) ^((2π)/3) (((arcsinhx)^2 )/(2cosx))dx

$\int_{\frac{π}{2}}^{\frac{2 π}{3}} \frac{{(arcsinhx)}^{2}}{2 cosx} dx$

Question Number 128025 Answers: 2 Comments: 0

Question Number 128023 Answers: 0 Comments: 0

....nice calculus...= Titu′s lemma:: for any positive numbers : a_1 ,a_2 ,...,a_n , b_1 ,b_2 ,...,b_n we have: (((a_1 +...+a_n )^2 )/(b_1 +...+b_n ))≤(a_1 ^2 /b_1 ) +...+(a_n ^2 /b_n ) proof : put : x=(x_1 ,...,x_n )∈R^n :y=(y_1 ,...,y_n )∈R^n (x.y)^2 ≤∣x∣^2 ∣y∣^2 (cauchy−schwarz inequality) (x_1 y_1 +...+x_n y_n )^2 ≤(x_(1 ) ^2 +...+x_n ^2 )(y_1 ^2 +...+y_(n ) ^2 ) by applying subsitution : x_i =(a_i /( (√b_i ))) , y_i =(√b_i ) (i=1,2 ,...,n) ((a_(1 ) ^2 +...+a_(n ) ^2 )/(b_2 +...+b_n ))≤(a_1 ^2 /b_1 )+...+(a_n ^2 /b_n ) ✓✓

$. . . . n i c e c a l c u l u s . . . =$ ${T i t u}^{'} s l e m m a ::$ $f o r a n y p o s i t i v e n u m b e r s :$ $a_{1}, a_{2}, . . ., a_{n}, b_{1}, b_{2}, . . ., b_{n}$ $w e h a v e :$ $\frac{{(a_{1} + . . . + a_{n})}^{2}}{b_{1} + . . . + b_{n}} ⩽ \frac{a_{1}^{2}}{b_{1}} + . . . + \frac{a_{n}^{2}}{b_{n}}$ $p r o o f :$ $p u t : x = (x_{1}, . . ., x_{n}) \in R^{n}$ $: y = (y_{1}, . . ., y_{n}) \in R^{n}$ ${(x . y)}^{2} ⩽∣ x ∣^{2} ∣ y ∣^{2} (c a u c h y - s c h w a r z i n e q u a l i t y)$ ${(x_{1} y_{1} + . . . + x_{n} y_{n})}^{2} ⩽ (x_{1}^{2} + . . . + x_{n}^{2}) (y_{1}^{2} + . . . + y_{n}^{2})$ $b y a p p l y i n g s u b s i t u t i o n :$ $x_{i} = \frac{a_{i}}{\sqrt{b_{i}}}, y_{i} = \sqrt{b_{i}} (i = 1, 2, . . ., n)$ $\frac{a_{1}^{2} + . . . + a_{n}^{2}}{b_{2} + . . . + b_{n}} ⩽ \frac{a_{1}^{2}}{b_{1}} + . . . + \frac{a_{n}^{2}}{b_{n}} ✓ ✓$

Question Number 127958 Answers: 1 Comments: 0

...nice calculus... calculate Ω=∫_1 ^( ∞) ((ln(x^4 −2x^2 +2))/(x(√(x^2 −1)) )) dx=?

$. . . n i c e c a l c u l u s . . .$ $c a l c u l a t e$ $Ω = \int_{1}^{\infty} \frac{l n (x^{4} - 2 x^{2} + 2)}{x \sqrt{x^{2} - 1}} d x = ?$

Question Number 127948 Answers: 1 Comments: 0

Question Number 127952 Answers: 1 Comments: 0

Question Number 127925 Answers: 1 Comments: 0

find F(a)=∫_0 ^1 (√((1+a^2 t^2 )/(1−t^2 ))) dt for background see Q127811.

$f i n d F (a) = \int_{0}^{1} \sqrt{\frac{1 + a^{2} t^{2}}{1 - t^{2}}} d t$ $f o r b a c k g r o u n d s e e Q 127811 .$

Question Number 127904 Answers: 2 Comments: 0

prove that ∫_0 ^( 100) (dx/( (√(x(100−x))))) = π

$prove that \int_{0}^{100} \frac{dx}{\sqrt{x (100 - x)}} = π$

Question Number 127885 Answers: 1 Comments: 0

∫(((sin (2tan^(−1) (x)+x))/x)) the limit [0,∞)

$\int (\frac{\sin ({2 \tan}^{- 1} (x) + x)}{x}) t h e l i m i t [0, \infty)$

Question Number 127870 Answers: 2 Comments: 0

2021 HAPPY NEW Year 1)∫((x^3 +3x+2)/((x^2 +1)^2 (x+1)))dx 2)∫((2cos(x)−sin(x))/(3sin(x)+5cos(x)))dx 3)∫((tan(2x))/( (√(sin^6 (x)+cos^6 (x)))))dx 4)∫x(√((1−x^2 )/(1+x^2 ))) dx

$2021$ $H A P P Y N E W Y e a r$ $1) \int \frac{x^{3} + 3 x + 2}{{(x^{2} + 1)}^{2} (x + 1)} d x$ $2) \int \frac{2 c o s (x) - s i n (x)}{3 s i n (x) + 5 c o s (x)} d x$ $3) \int \frac{t a n (2 x)}{\sqrt{{s i n}^{6} (x) + {c o s}^{6} (x)}} d x$ $4) \int x \sqrt{\frac{1 - x^{2}}{1 + x^{2}}} d x$