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AllQuestion and Answers: Page 1562

Question Number 53358 Answers: 1 Comments: 0

∫ (1/(√(sin^3 x cos x))) dx =

$$\int\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{sin}^{\mathrm{3}} {x}\:\mathrm{cos}\:{x}}}\:{dx}\:= \\ $$

Question Number 53357 Answers: 1 Comments: 0

∫ (((x^4 −x)^(1/4) )/x^5 ) dx =

$$\int\:\:\frac{\left({x}^{\mathrm{4}} −{x}\right)^{\mathrm{1}/\mathrm{4}} }{{x}^{\mathrm{5}} }\:{dx}\:= \\ $$

Question Number 53353 Answers: 1 Comments: 0

Question Number 53351 Answers: 0 Comments: 1

Question Number 53349 Answers: 0 Comments: 2

Question Number 53342 Answers: 0 Comments: 4

Question Number 53340 Answers: 1 Comments: 0

Question Number 53330 Answers: 0 Comments: 12

Question Number 53328 Answers: 1 Comments: 2

Question Number 53325 Answers: 0 Comments: 10

Question Number 53324 Answers: 1 Comments: 0

If 4a + 5b + 9c=36 and 7a + 9b + 17c=66, then a+b+c=_____.

$$\mathrm{If}\:\mathrm{4}{a}\:+\:\mathrm{5}{b}\:+\:\mathrm{9}{c}=\mathrm{36}\:\mathrm{and}\:\mathrm{7}{a}\:+\:\mathrm{9}{b}\:+\:\mathrm{17}{c}=\mathrm{66}, \\ $$$$\mathrm{then}\:{a}+{b}+{c}=\_\_\_\_\_. \\ $$

Question Number 53323 Answers: 1 Comments: 0

Question Number 53318 Answers: 1 Comments: 1

Question Number 53311 Answers: 1 Comments: 1

If ∫ ((4e^x +6e^(−x) )/(9e^x −4e^(−x) )) dx=Ax+B log(9e^(2x) −4)+C then A=... B=... C=...

$$\mathrm{If}\:\int\:\frac{\mathrm{4}{e}^{{x}} +\mathrm{6}{e}^{−{x}} }{\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} }\:{dx}={Ax}+{B}\:\mathrm{log}\left(\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}\right)+{C} \\ $$$$\mathrm{then} \\ $$$${A}=... \\ $$$${B}=... \\ $$$${C}=... \\ $$

Question Number 53295 Answers: 1 Comments: 1

∫_0 ^(π/2) (1/(2+cos x)) dx=...

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}+\mathrm{cos}\:{x}}\:{dx}=... \\ $$

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=...

$$\int_{−\mathrm{1}/\mathrm{2}} ^{\mathrm{1}/\mathrm{2}} \left[\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{2}\right]^{\mathrm{1}/\mathrm{2}} {dx}=... \\ $$

Question Number 53293 Answers: 1 Comments: 1

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

$$\int_{−\mathrm{1}/\mathrm{2}} ^{\mathrm{1}/\mathrm{2}} \mid{x}\mathrm{cos}\:\frac{\pi{x}}{\mathrm{2}}\mid\:{dx}=... \\ $$

Question Number 53292 Answers: 1 Comments: 1

∫_0 ^1 e^x^2 dx=..

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

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.

$${let}\:{I}_{\lambda} \:=\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{xdx}}{{cos}^{\mathrm{2}} {x}\:+\lambda^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}\:\:{with}\:\lambda\:{real} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{value}\:{of}\:{I}_{\lambda} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{xdx}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}\:+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}\:{with}\:{a}\:{and}\:{b}\:{reals}. \\ $$

Question Number 53284 Answers: 0 Comments: 2

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

$${find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({xt}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\:\:{with}\:{x}\:{real}\:. \\ $$

Question Number 53277 Answers: 1 Comments: 1

∫_( 0) ^1 (x/((1−x)^(3/4) )) dx =

$$\underset{\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\:\:\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}/\mathrm{4}} }\:{dx}\:= \\ $$

Question Number 53276 Answers: 1 Comments: 1

Question Number 53273 Answers: 1 Comments: 1

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 .

$$\left.\mathrm{1}\right){calculate}\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{xt}^{\mathrm{2}} } {dt}\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}^{\mathrm{2}} } \:−{e}^{−\mathrm{2}{t}^{\mathrm{2}} } }{{t}^{\mathrm{2}} }\:{dt}\:\:{by}\:{using} \\ $$$${fubinni}\:{theorem}\:. \\ $$

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 .

$$\left.\mathrm{1}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{at}} {dt}\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right){by}\:{using}\:{fubinni}\:{theorem}\:{find}\:{the}\:{value}\:{of} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} \:−{e}^{−{xt}} }{{t}}{dt}\:\:\:{with}\:{x}>\mathrm{0}\:. \\ $$

Question Number 53262 Answers: 1 Comments: 2

find x: (1/(√(x + 1 + (√x)))) − (2/(√(x − 2 + (√x)))) = (√(x − 1))

$$\mathrm{find}\:\mathrm{x}:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{x}\:+\:\mathrm{1}\:+\:\sqrt{\mathrm{x}}}}\:\:−\:\:\frac{\mathrm{2}}{\sqrt{\mathrm{x}\:−\:\mathrm{2}\:+\:\sqrt{\mathrm{x}}}}\:\:=\:\:\sqrt{\mathrm{x}\:−\:\mathrm{1}} \\ $$

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