Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 1570

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))) .

$${calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{1}+\mathrm{2}{sinx}}{\mathrm{3}\:+\mathrm{2}{cosx}}{dx} \\ $$$${let}\:{A}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{1}+\mathrm{2}{sinx}}{\mathrm{3}\:+\mathrm{2}{cosx}}{dx}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}+\frac{\mathrm{4}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{3}+\mathrm{2}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\frac{\mathrm{3}+\mathrm{3}{t}^{\mathrm{2}} +\mathrm{2}−\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)}{dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\mathrm{2}} +\mathrm{4}{t}\:+\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{5}+{t}^{\mathrm{2}} \right)}{dt}\:\:{let}\:{decompose}\:{F}\left({t}\right)=\frac{{t}^{\mathrm{2}} \:+\mathrm{4}{t}+\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{5}\right)} \\ $$$${F}\left({t}\right)=\frac{{at}\:+{b}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{5}}\:\Rightarrow\left({at}+{b}\right)\left({t}^{\mathrm{2}} \:+\mathrm{5}\right)+\left({ct}+{d}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\:={t}^{\mathrm{2}} \:+\mathrm{4}{t}\:+\mathrm{1}\:\Rightarrow \\ $$$${at}^{\mathrm{3}} \:+\mathrm{5}{at}\:+{bt}^{\mathrm{2}} \:+\mathrm{5}{b}\:+{ct}^{\mathrm{3}} \:+{ct}\:+{dt}^{\mathrm{2}} \:+{d}\:={t}^{\mathrm{2}} \:+\mathrm{4}{t}\:+\mathrm{1}\:\Rightarrow \\ $$$$\left({a}+{c}\right){t}^{\mathrm{3}} \:+\left({b}+{d}\right){t}^{\mathrm{2}} \:+\left(\mathrm{5}{a}+{c}\right){t}\:+\mathrm{5}{b}\:+{d}\:={t}^{\mathrm{2}} \:+\mathrm{4}{t}\:+\mathrm{1}\:\Rightarrow{a}+{c}=\mathrm{0}\:{and}\:{b}+{d}=\mathrm{1}\:{and} \\ $$$$\mathrm{5}{a}+{c}\:=\mathrm{4}\:{and}\:\mathrm{5}{b}+{d}\:=\mathrm{1}\:\Rightarrow{c}=−{a}\:\Rightarrow{a}=\mathrm{1}\:\Rightarrow{c}=−\mathrm{1}\: \\ $$$${we}\:{have}\:{d}=\mathrm{1}−{b}\:\Rightarrow\mathrm{5}{b}\:+\mathrm{1}−{b}\:=\mathrm{1}\:\Rightarrow{b}=\mathrm{0}\:\Rightarrow{d}=\mathrm{1}\:\Rightarrow \\ $$$${F}\left({t}\right)=\frac{{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{t}\:+\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{5}}\:\:\Rightarrow\:{A}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:{F}\left({t}\right){dt}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:+\int_{\mathrm{0}} ^{\infty} \:\frac{−\mathrm{2}{t}\:+\mathrm{2}}{{t}^{\mathrm{2}} \:+\mathrm{5}}{dt} \\ $$$$=\left[{ln}\left(\frac{{t}^{\mathrm{2}} \:+\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{5}}\right)\right]_{\mathrm{0}} ^{+\infty} \:\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{5}}\:={ln}\left(\mathrm{5}\right)\:+\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{5}}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{5}}{dt}\:=_{{t}\:=\sqrt{\mathrm{5}}{u}\:} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\sqrt{\mathrm{5}}{du}}{\mathrm{5}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\:\left[{artanu}\right]_{\mathrm{0}} ^{+\infty} \:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{5}}}\:\Rightarrow \\ $$$${A}\:={ln}\left(\mathrm{5}\right)\:+\frac{\pi}{\mathrm{2}\sqrt{\mathrm{5}}}\:. \\ $$

Question Number 53108    Answers: 1   Comments: 0

((x + 2))^(1/3) − ((x − 3))^(1/3) > (1/2)

$$\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:+\:\mathrm{2}}\:\:−\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:−\:\mathrm{3}}\:\:\:>\:\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Question Number 53086    Answers: 0   Comments: 1

JEE Main chemistry errors by Resonance. Today is the last day for raising objectiions. You need to tap once to select and tap the text again to open hyperlink.

$$\mathrm{JEE}\:\mathrm{Main}\:\mathrm{chemistry}\:\mathrm{errors}\:\mathrm{by} \\ $$$$\mathrm{Resonance}.\:\mathrm{Today}\:\mathrm{is}\:\mathrm{the}\:\mathrm{last}\:\mathrm{day} \\ $$$$\mathrm{for}\:\mathrm{raising}\:\mathrm{objectiions}. \\ $$$$\mathrm{You}\:\mathrm{need}\:\mathrm{to}\:\mathrm{tap}\:\mathrm{once}\:\mathrm{to}\:\mathrm{select}\:\mathrm{and}\:\mathrm{tap} \\ $$$$\mathrm{the}\:\mathrm{text}\:\mathrm{again}\:\mathrm{to}\:\mathrm{open}\:\mathrm{hyperlink}. \\ $$

Question Number 53081    Answers: 3   Comments: 0

Question Number 53080    Answers: 1   Comments: 1

calculate ∫_0 ^π ((cos^2 x)/(2+3sin(2x)))dx

$${calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{cos}^{\mathrm{2}} {x}}{\mathrm{2}+\mathrm{3}{sin}\left(\mathrm{2}{x}\right)}{dx} \\ $$

Question Number 53078    Answers: 1   Comments: 1

∫_0 ^1 (1/((x^3 +1)^(3/2) )) dx=...

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }\:{dx}=... \\ $$

Question Number 53071    Answers: 0   Comments: 6

Question Number 53066    Answers: 1   Comments: 9

Question Number 53089    Answers: 0   Comments: 0

∫_( (π/2) ) ^( ∞) (dx/((5 + x^2 ) tanh^(−1) ((x/3))))

$$\int_{\:\frac{\pi}{\mathrm{2}}\:} ^{\:\infty} \:\:\:\frac{\mathrm{dx}}{\left(\mathrm{5}\:+\:\mathrm{x}^{\mathrm{2}} \right)\:\mathrm{tanh}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{3}}\right)} \\ $$

Question Number 53055    Answers: 0   Comments: 4

The area of △XYZ is 50cm^2 . L is the midpoint of XZ andKY ∥LM. find the area of △ZLY. help please sir

$${The}\:{area}\:{of}\:\bigtriangleup{XYZ}\:{is}\:\mathrm{50}{cm}^{\mathrm{2}} .\:{L}\: \\ $$$${is}\:{the}\:{midpoint}\:{of}\:{XZ}\:{andKY}\:\parallel{LM}. \\ $$$${find}\:{the}\:{area}\:{of}\:\:\bigtriangleup{ZLY}. \\ $$$${help}\:{please}\:{sir} \\ $$

Question Number 53063    Answers: 1   Comments: 0

Comment resoudre cette eqution 2^x ×3−y^2 =−1

$$\mathrm{Comment}\:\mathrm{resoudre}\:\mathrm{cette}\:\mathrm{eqution}\: \\ $$$$\mathrm{2}^{\mathrm{x}} ×\mathrm{3}−\mathrm{y}^{\mathrm{2}} =−\mathrm{1} \\ $$

Question Number 53051    Answers: 2   Comments: 0

∫sin (2x)cos xd(x)=

$$\int\mathrm{sin}\:\left(\mathrm{2}{x}\right)\mathrm{cos}\:{xd}\left({x}\right)= \\ $$

Question Number 53050    Answers: 0   Comments: 0

$$ \\ $$

Question Number 53048    Answers: 0   Comments: 1

Question Number 53043    Answers: 2   Comments: 1

Question Number 53034    Answers: 2   Comments: 0

Angle between the lines ((x−1)/1)=((y−1)/1)=((z−1)/2)and ((x−1)/(−(√(3−1))))=((y−1)/(√(3−1)))=((z−1)/4) is

$$\mathrm{Angle}\:\mathrm{between}\:\mathrm{the}\:\mathrm{lines}\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{1}}=\frac{\mathrm{y}−\mathrm{1}}{\mathrm{1}}=\frac{\mathrm{z}−\mathrm{1}}{\mathrm{2}}\mathrm{and}\:\frac{\mathrm{x}−\mathrm{1}}{−\sqrt{\mathrm{3}−\mathrm{1}}}=\frac{\mathrm{y}−\mathrm{1}}{\sqrt{\mathrm{3}−\mathrm{1}}}=\frac{\mathrm{z}−\mathrm{1}}{\mathrm{4}}\:\mathrm{is} \\ $$

Question Number 53033    Answers: 0   Comments: 0

in the law of mean the value of θ satisfies the condition

$$\mathrm{in}\:\mathrm{the}\:\mathrm{law}\:\mathrm{of}\:\mathrm{mean}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\theta\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{condition}\: \\ $$

Question Number 53031    Answers: 1   Comments: 1

Question Number 53025    Answers: 1   Comments: 1

Question Number 53007    Answers: 0   Comments: 0

2^x ×3−y^2 =−1

$$\mathrm{2}^{\mathrm{x}} ×\mathrm{3}−\mathrm{y}^{\mathrm{2}} =−\mathrm{1} \\ $$

Question Number 52999    Answers: 0   Comments: 6

∫_0 ^( ∞) ((x ln^2 (x))/(e^x − 1)) dx

$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{1}}\:\:\boldsymbol{\mathrm{dx}}\:\:\: \\ $$

Question Number 52993    Answers: 0   Comments: 2

Question Number 52992    Answers: 3   Comments: 1

Question Number 52991    Answers: 1   Comments: 1

Question Number 52988    Answers: 1   Comments: 0

∫ (x^2 /(√(1 + x^4 ))) dx

$$\int\:\frac{{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{4}} }}\:{dx} \\ $$

Question Number 52960    Answers: 1   Comments: 0

A wholesaler buys an article at“ $32 less 12.5%”. she then sells the article at a gain of 25% of her cost after allowing a 20% discount on her marked price. what is the marked price. please sir help me thanks

$${A}\:{wholesaler}\:{buys}\:{an}\:{article}\:{at}``\:\$\mathrm{32}\: \\ $$$${less}\:\mathrm{12}.\mathrm{5\%}''.\:{she}\:{then}\:{sells}\:{the}\:{article} \\ $$$${at}\:{a}\:{gain}\:{of}\:\mathrm{25\%}\:{of}\:{her}\:{cost}\:{after}\: \\ $$$${allowing}\:{a}\:\:\mathrm{20\%}\:{discount}\:{on}\:{her}\: \\ $$$${marked}\:{price}.\:{what}\:{is}\:{the}\:{marked}\: \\ $$$${price}. \\ $$$${please}\:{sir}\:{help}\:{me}\:{thanks} \\ $$

  Pg 1565      Pg 1566      Pg 1567      Pg 1568      Pg 1569      Pg 1570      Pg 1571      Pg 1572      Pg 1573      Pg 1574   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com