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Question Number 86461    Answers: 3   Comments: 0

Use exponential representation of sin θ and cos θ to show that a) sin^2 θ + cos^2 θ = 1 b) cos^2 θ − sin^2 θ = cos2θ c) 2 sinθ cosθ = 2sin2θ.

$$\mathrm{Use}\:\mathrm{exponential}\:\mathrm{representation}\:\mathrm{of}\:\mathrm{sin}\:\theta\:\mathrm{and}\:\mathrm{cos}\:\theta\:\mathrm{to}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left.\mathrm{a}\left.\right)\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\:\mathrm{cos}^{\mathrm{2}} \:\theta\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\right)\:\mathrm{cos}^{\mathrm{2}} \theta\:−\:\mathrm{sin}^{\mathrm{2}} \theta\:=\:\mathrm{cos2}\theta \\ $$$$\left.\mathrm{c}\right)\:\mathrm{2}\:\mathrm{sin}\theta\:\mathrm{cos}\theta\:=\:\mathrm{2sin2}\theta. \\ $$

Question Number 86365    Answers: 0   Comments: 0

I think it will be ∫_0 ^(π/4) (dx/(√(1+tanx))) ≈∫_0 ^(π/4) (dx/(√(1+x))) =(𝛑/4)−(1/2).(1/2).((𝛑/4))^2 +((1.3)/(2.4)).(1/3).((𝛑/4))^3 −((1.3.5)/(2.4.6)).(1/4)((𝛑/4))^4 +....

$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{dx}}{\sqrt{\mathrm{1}+\mathrm{tanx}}}\:\approx\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{dx}}{\sqrt{\mathrm{1}+\mathrm{x}}}\: \\ $$$$=\frac{\boldsymbol{\pi}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}.\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}}.\frac{\mathrm{1}}{\mathrm{3}}.\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)^{\mathrm{3}} −\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}.\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)^{\mathrm{4}} +.... \\ $$

Question Number 86356    Answers: 1   Comments: 1

Question Number 86198    Answers: 0   Comments: 7

any methods to sketch these curves r = a(1−cosθ) r= a + b cosθ a>b r= a + bcosθ a<b

$$\mathrm{any}\:\mathrm{methods}\:\mathrm{to}\:\mathrm{sketch}\:\mathrm{these}\:\mathrm{curves} \\ $$$$\:\mathrm{r}\:=\:\mathrm{a}\left(\mathrm{1}−\mathrm{cos}\theta\right) \\ $$$$\:\mathrm{r}=\:\mathrm{a}\:+\:\mathrm{b}\:\mathrm{cos}\theta\:\:{a}>{b} \\ $$$$\:\mathrm{r}=\:\mathrm{a}\:+\:\mathrm{bcos}\theta\:\:\:{a}<{b} \\ $$

Question Number 86140    Answers: 0   Comments: 1

Question Number 86132    Answers: 1   Comments: 0

∫x^3 sin(2x^2 +6)^5 dx

$$\int{x}^{\mathrm{3}} \:{sin}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}\right)^{\mathrm{5}} \:{dx} \\ $$

Question Number 86128    Answers: 1   Comments: 0

⌊2x−(1/2)⌋=⌊∣x∣−(1/2)⌋=2x−2

$$\lfloor\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}\rfloor=\lfloor\mid{x}\mid−\frac{\mathrm{1}}{\mathrm{2}}\rfloor=\mathrm{2}{x}−\mathrm{2} \\ $$

Question Number 86116    Answers: 0   Comments: 1

lim_(x→∞) ((tanx)/x)

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{tanx}}{{x}} \\ $$

Question Number 86062    Answers: 2   Comments: 1

∫((√(x^2 −25))/x)dx

$$\int\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{25}}}{{x}}{dx} \\ $$

Question Number 85982    Answers: 0   Comments: 2

A primitive of the function defned by f(x) = x −1 + (1/(x+1)) is A. F(x) = (x^2 /2) −x + ln(x + 1) B. F(x) = (x^2 /2) + ln(x−1) C. F(x) = (x^2 /2)−x + ln(1−x) D. F(x) = −x + ln(x−1)

$$\mathrm{A}\:\mathrm{primitive}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{defned}\:\mathrm{by}\:\mathrm{f}\left({x}\right)\:=\:{x}\:−\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\mathrm{is}\: \\ $$$$\mathrm{A}.\:\mathrm{F}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:−{x}\:+\:\mathrm{ln}\left({x}\:+\:\mathrm{1}\right)\:\:\:\:\mathrm{B}.\:\mathrm{F}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\:\mathrm{ln}\left({x}−\mathrm{1}\right) \\ $$$$\mathrm{C}.\:\mathrm{F}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\:+\:\mathrm{ln}\left(\mathrm{1}−{x}\right)\:\:\:\:\:\:\:\:\:\mathrm{D}.\:\mathrm{F}\left({x}\right)\:=\:−{x}\:+\:\mathrm{ln}\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$

Question Number 85888    Answers: 0   Comments: 5

i−∫_0 ^5 (x+[2x])^([(x/3)]) dx ii−∫_0 ^3 (z−{z})^([z]) dz

$${i}−\int_{\mathrm{0}} ^{\mathrm{5}} \left({x}+\left[\mathrm{2}{x}\right]\right)^{\left[\frac{{x}}{\mathrm{3}}\right]} {dx} \\ $$$$ \\ $$$${ii}−\int_{\mathrm{0}} ^{\mathrm{3}} \left({z}−\left\{{z}\right\}\right)^{\left[{z}\right]} \:{dz} \\ $$

Question Number 85872    Answers: 1   Comments: 1

∫cos^(2020) x dx = ?

$$\int\mathrm{cos}^{\mathrm{2020}} \mathrm{x}\:\mathrm{dx}\:=\:? \\ $$

Question Number 85858    Answers: 0   Comments: 0

Is a matrix A^T A always positive definite?

$$\mathrm{Is}\:\mathrm{a}\:\mathrm{matrix} \\ $$$$\mathrm{A}^{\mathrm{T}} \mathrm{A}\:\mathrm{always}\:\mathrm{positive}\:\mathrm{definite}? \\ $$

Question Number 85698    Answers: 0   Comments: 0

please any recommendation of a youtube video on General conics??

$$\mathrm{please}\:\mathrm{any}\:\mathrm{recommendation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{youtube}\:\mathrm{video} \\ $$$$\mathrm{on}\:\mathrm{General}\:\mathrm{conics}?? \\ $$

Question Number 85676    Answers: 0   Comments: 15

∫ _0 ^∞ (dx/((x+(√(1+x^2 )))^2 )) let x = tan t ⇒dx=sec^2 t dt ∫_0 ^(π/2) ((sec^2 t dt)/((tan t+sec t)^2 )) = ∫_0 ^(π/2) (dt/((sin t+1)^2 )) = ∫_0 ^(π/2) (dt/((cos (1/2)t+sin (1/2)t)^4 )) = ∫_0 ^(π/2) (dt/(4cos^4 ((1/2)t−(π/4)))) = (1/4)∫_0 ^(π/2) sec^4 ((1/2)t−(π/4)) dt [ let (1/2)t−(π/4)= u] = (1/4)∫_(−(π/4)) ^0 sec^4 u ×2du =(1/2)∫ _(−(π/4)) ^0 (tan^2 u+1) d(tan u) = (1/2) [(1/3)tan^3 u + tan u ]_(−(π/4)) ^0 = (1/2) [ 0−(−(1/3)−1)]= (2/3)

$$\int\underset{\mathrm{0}} {\overset{\infty} {\:}}\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$${let}\:{x}\:=\:\mathrm{tan}\:{t}\:\Rightarrow{dx}=\mathrm{sec}\:^{\mathrm{2}} {t}\:{dt} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} {t}\:{dt}}{\left(\mathrm{tan}\:{t}+\mathrm{sec}\:{t}\right)^{\mathrm{2}} }\:=\: \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\left(\mathrm{sin}\:{t}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\left(\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{t}+\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{t}\right)^{\mathrm{4}} } \\ $$$$=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\mathrm{4cos}^{\mathrm{4}} \:\left(\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{sec}\:^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}\right)\:{dt} \\ $$$$\left[\:{let}\:\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}=\:{u}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\mathrm{0}} {\int}}\:\mathrm{sec}\:^{\mathrm{4}} {u}\:×\mathrm{2}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\mathrm{0}} {\:}}\left(\mathrm{tan}\:^{\mathrm{2}} {u}+\mathrm{1}\right)\:{d}\left(\mathrm{tan}\:{u}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{3}} {u}\:+\:\mathrm{tan}\:{u}\:\right]_{−\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:\mathrm{0}−\left(−\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}\right)\right]=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$

Question Number 85580    Answers: 0   Comments: 0

Solve: (D^2 +2D+1)y= x cos x

$$\boldsymbol{\mathrm{Solve}}: \\ $$$$\:\left(\mathrm{D}^{\mathrm{2}} +\mathrm{2D}+\mathrm{1}\right)\mathrm{y}=\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x} \\ $$$$ \\ $$

Question Number 85354    Answers: 0   Comments: 0

∫x(sin(cos(x)))^(x−1) dx

$$\int{x}\left({sin}\left({cos}\left({x}\right)\right)\right)^{{x}−\mathrm{1}} \:\:{dx} \\ $$

Question Number 85286    Answers: 0   Comments: 2

Question Number 85279    Answers: 3   Comments: 0

who can help write out the mechanism for the reaction C_6 H_6 →_(HNO_3 ) ^(H_2 SO_4 ) C_6 H_5 NO_2

$$\mathrm{who}\:\mathrm{can}\:\mathrm{help}\:\mathrm{write}\:\mathrm{out}\:\mathrm{the}\:\mathrm{mechanism}\:\mathrm{for}\:\mathrm{the}\: \\ $$$$\mathrm{reaction} \\ $$$$\:\:\:\mathrm{C}_{\mathrm{6}} \mathrm{H}_{\mathrm{6}} \:\underset{\mathrm{HNO}_{\mathrm{3}} } {\overset{\mathrm{H}_{\mathrm{2}} \mathrm{SO}_{\mathrm{4}} } {\rightarrow}}\:\:\mathrm{C}_{\mathrm{6}} \mathrm{H}_{\mathrm{5}} \mathrm{NO}_{\mathrm{2}} \\ $$

Question Number 85262    Answers: 1   Comments: 6

find the centre of symmetry of the curve y = (1/(x + 2))

$$\mathrm{find}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{symmetry}\:\mathrm{of}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\:\:\:{y}\:=\:\frac{\mathrm{1}}{{x}\:+\:\mathrm{2}} \\ $$

Question Number 85222    Answers: 0   Comments: 3

In C++ What is the sections doing ? and What is the output from the sections below ? a. int p=o; for (int i=1; i<=30; i++){ if (i%5==0) p+=i; } cout<<p<<endl; b. int count=2; int g=0; while (count<=50){ g=count; cout<<g<<endl; count +=2; } cout<<count<<endl;

$$\mathrm{In}\:\mathrm{C}++ \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sections}\:\mathrm{doing}\:?\:\mathrm{and} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{output}\:\mathrm{from}\:\mathrm{the}\:\mathrm{sections}\: \\ $$$$\mathrm{below}\:? \\ $$$$\mathrm{a}.\:\mathrm{int}\:\mathrm{p}=\mathrm{o}; \\ $$$$\:\:\:\:\:\mathrm{for}\:\left(\mathrm{int}\:\mathrm{i}=\mathrm{1};\:\mathrm{i}<=\mathrm{30};\:\mathrm{i}++\right)\left\{\right. \\ $$$$\:\:\:\:\:\mathrm{if}\:\left(\mathrm{i\%5}==\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\mathrm{p}+=\mathrm{i};\:\: \\ $$$$\left.\:\:\:\:\:\:\:\right\} \\ $$$$\:\:\:\:\:\:\:\mathrm{cout}<<\mathrm{p}<<\mathrm{endl}; \\ $$$$ \\ $$$$ \\ $$$$\mathrm{b}.\:\:\mathrm{int}\:\mathrm{count}=\mathrm{2}; \\ $$$$\:\:\:\:\:\:\mathrm{int}\:\mathrm{g}=\mathrm{0}; \\ $$$$\:\:\:\:\:\:\mathrm{while}\:\left(\mathrm{count}<=\mathrm{50}\right)\left\{\right. \\ $$$$\:\:\:\:\:\:\mathrm{g}=\mathrm{count}; \\ $$$$\:\:\:\:\:\:\mathrm{cout}<<\mathrm{g}<<\mathrm{endl}; \\ $$$$\:\:\:\:\:\:\mathrm{count}\:+=\mathrm{2}; \\ $$$$\left.\:\:\:\:\:\:\:\right\} \\ $$$$\:\:\:\:\:\:\:\mathrm{cout}<<\mathrm{count}<<\mathrm{endl}; \\ $$

Question Number 85142    Answers: 1   Comments: 0

show that ∫_0 ^n [x^2 ]dx =n(n^2 −1)−Σ_(k=1) ^(n^2 −1) (√k)

$${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{{n}} \left[{x}^{\mathrm{2}} \right]{dx}\:={n}\left({n}^{\mathrm{2}} −\mathrm{1}\right)−\underset{{k}=\mathrm{1}} {\overset{{n}^{\mathrm{2}} −\mathrm{1}} {\sum}}\sqrt{{k}}\: \\ $$

Question Number 85131    Answers: 0   Comments: 4

what procedure will you use to find the inverse of A = ((2,1,9),(1,5,1),(3,0,3) )

$$\mathrm{what}\:\mathrm{procedure}\:\mathrm{will}\:\mathrm{you}\:\mathrm{use}\:\mathrm{to}\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{of} \\ $$$$\:\mathrm{A}\:=\:\begin{pmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{9}}\\{\mathrm{1}}&{\mathrm{5}}&{\mathrm{1}}\\{\mathrm{3}}&{\mathrm{0}}&{\mathrm{3}}\end{pmatrix} \\ $$

Question Number 85129    Answers: 0   Comments: 2

lim_(x→e) [∫_0 ^e ((1/x))dx] =?

$$\underset{{x}\rightarrow{e}} {\mathrm{lim}}\:\left[\underset{\mathrm{0}} {\overset{{e}} {\int}}\left(\frac{\mathrm{1}}{{x}}\right){dx}\right]\:=? \\ $$

Question Number 85127    Answers: 1   Comments: 4

evaluate: lim_(x→0) (√x) ln(sin x)

$$\mathrm{evaluate}: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt{{x}}\:\mathrm{ln}\left(\mathrm{sin}\:{x}\right) \\ $$$$ \\ $$

Question Number 85083    Answers: 0   Comments: 1

a^3 −b^3 =...?

$${a}^{\mathrm{3}} −{b}^{\mathrm{3}} =...? \\ $$

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