Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 1271

Question Number 80053    Answers: 0   Comments: 4

Find integer x, y such that 2^x −y^2 =615

$${Find}\:{integer}\:{x},\:{y}\:{such}\:{that} \\ $$$$\mathrm{2}^{{x}} −{y}^{\mathrm{2}} =\mathrm{615} \\ $$

Question Number 80052    Answers: 0   Comments: 0

∫ e^(sin 2x) .cos x dx =

$$\int\:\mathrm{e}^{\mathrm{sin}\:\mathrm{2x}} .\mathrm{cos}\:\mathrm{x}\:\mathrm{dx}\:= \\ $$$$ \\ $$

Question Number 80108    Answers: 1   Comments: 3

a,b,c ∈R ((b+c+d)/a)=((a+c+d)/b)=((a+b+c)/d)=((a+b+d)/c)=r what is r?

$${a},{b},{c}\:\in\mathbb{R} \\ $$$$\frac{{b}+{c}+{d}}{{a}}=\frac{{a}+{c}+{d}}{{b}}=\frac{{a}+{b}+{c}}{{d}}=\frac{{a}+{b}+{d}}{{c}}={r} \\ $$$${what}\:{is}\:{r}? \\ $$

Question Number 80039    Answers: 1   Comments: 6

prove that (1+x)(1+(1/x))≥4

$${prove}\:{that} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\geqslant\mathrm{4} \\ $$

Question Number 80037    Answers: 0   Comments: 0

A matrix A= [(a_(ij) ) ] is an upper triangular matrix if

$$\mathrm{A}\:\mathrm{matrix}\:{A}=\begin{bmatrix}{{a}_{{ij}} }\end{bmatrix}\:\mathrm{is}\:\mathrm{an}\:\mathrm{upper}\:\mathrm{triangular} \\ $$$$\mathrm{matrix}\:\mathrm{if} \\ $$

Question Number 80036    Answers: 1   Comments: 3

Σ_(n=1) ^∞ (1/((n+1)(n+2)(n+3)))=

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}=\: \\ $$

Question Number 80027    Answers: 0   Comments: 4

find minimum value of (√(x^2 +4))+(√(x^2 −24x+153)) for x≥0 in R

$${find}\:{minimum} \\ $$$${value}\:{of}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{153}} \\ $$$${for}\:{x}\geqslant\mathrm{0}\:{in}\:\mathbb{R} \\ $$

Question Number 80015    Answers: 2   Comments: 2

Question Number 79992    Answers: 0   Comments: 2

lim_(x→0) [(1/x)] = ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\:=\:? \\ $$

Question Number 80000    Answers: 1   Comments: 2

prove that lim_(x→0) ((arcsin(x/(√(1−x^2 ))))/(ln(1−x))) = −1

$$\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$$$\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{arcsin}}\frac{\boldsymbol{{x}}}{\sqrt{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }}}{\boldsymbol{{ln}}\left(\mathrm{1}−\boldsymbol{{x}}\right)}\:=\:−\mathrm{1} \\ $$

Question Number 79998    Answers: 0   Comments: 3

given 3x + 4y+1 = 3(√x) + 2(√y) find the value of (√(x.y))

$$\mathrm{given}\:\mathrm{3x}\:+\:\mathrm{4y}+\mathrm{1}\:=\:\mathrm{3}\sqrt{\mathrm{x}}\:+\:\mathrm{2}\sqrt{\mathrm{y}}\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\sqrt{\mathrm{x}.\mathrm{y}}\: \\ $$

Question Number 79978    Answers: 3   Comments: 0

Given for x,y,z>0: 2^x =3^y =5^z Arrange 2x, 3y, 5z in increasing order.

$${Given}\:{for}\:{x},{y},{z}>\mathrm{0}: \\ $$$$\mathrm{2}^{{x}} =\mathrm{3}^{{y}} =\mathrm{5}^{{z}} \\ $$$${Arrange}\:\mathrm{2}{x},\:\mathrm{3}{y},\:\mathrm{5}{z}\:{in}\:{increasing}\:{order}. \\ $$

Question Number 79974    Answers: 1   Comments: 0

Question Number 80004    Answers: 1   Comments: 2

x and y any integer satisfy equation (x−2004)(x−2006)=2^y the greatest possible value of x+y

$${x}\:{and}\:{y}\:{any}\:{integer}\:{satisfy} \\ $$$${equation}\:\left({x}−\mathrm{2004}\right)\left({x}−\mathrm{2006}\right)=\mathrm{2}^{{y}} \\ $$$${the}\:{greatest}\:{possible}\:{value} \\ $$$${of}\:{x}+{y} \\ $$

Question Number 79969    Answers: 1   Comments: 0

find the general solution for 2sin 3x = sin 2x

$${find}\:{the}\:{general}\:{solution}\:{for}\: \\ $$$$\:\:\mathrm{2sin}\:\mathrm{3}{x}\:=\:\mathrm{sin}\:\mathrm{2}{x} \\ $$

Question Number 79968    Answers: 0   Comments: 3

Find the 50^(th) entry of 3.127356432...

$${Find}\:{the}\:\mathrm{50}^{{th}} \:{entry}\:{of}\:\:\mathrm{3}.\mathrm{127356432}... \\ $$

Question Number 79966    Answers: 1   Comments: 1

Question Number 79950    Answers: 0   Comments: 3

If ∫_a ^b (x^n /(x^n +(16−x)^n )) dx = 6, then

$$\mathrm{If}\:\underset{{a}} {\overset{{b}} {\int}}\:\:\frac{{x}^{{n}} }{{x}^{{n}} +\left(\mathrm{16}−{x}\right)^{{n}} }\:{dx}\:=\:\mathrm{6},\:\mathrm{then}\: \\ $$

Question Number 79932    Answers: 1   Comments: 1

Question Number 79929    Answers: 0   Comments: 0

∫e^(√(sin x)) dx=?

$$\int{e}^{\sqrt{\mathrm{sin}\:{x}}} {dx}=? \\ $$

Question Number 79913    Answers: 0   Comments: 1

Convergence of I=∫_0 ^( ∞) (e^t /(e^(−t) +e^(2t) ∣sint∣))dt

$$\:{Convergence}\:\:{of}\:\:{I}=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{{t}} }{{e}^{−{t}} +{e}^{\mathrm{2}{t}} \mid{sint}\mid}{dt} \\ $$

Question Number 79903    Answers: 1   Comments: 11

Question Number 79883    Answers: 0   Comments: 9

to Sir Jagoll (and of course everybody else) (1) y=((x^2 −x−6)/(x^2 −3x−4))= =(((x−3)(x+2))/((x−4)(x+1))) ⇒ ⇒ { ((zeros at x=−2; x=3)),((vertical asymptotes at x=−1; x=4)) :} defined for x∈R\{−1; 4} range: transforming y=((x^2 −x−6)/(x^2 −3x−4)) to x^2 −((3y−1)/(y−1))x−((2(2y−3))/(y−1))=0 D=((25y^2 −46y+25)/(4(y−1)^2 ))>0∀x∈R ⇒ ⇒ { ((range=R)),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=1 ⇒ y=1 y′=−((2(x^2 −2x+7))/((x−4)^2 (x+1)^2 )) no real zeros ⇒ ⇒ no local extremes y′′=((4(x^3 −3x^2 +21x−25))/((x−4)^3 (x+1)^3 ))=0 at x≈1.33131 ⇒ ⇒ turning point (2) y=((x^2 −x−6)/(x^2 −3x+4))= =(((x−3)(x+2))/(x^2 −3x+4)); x^2 −3x+4=0 no real zeros ⇒ ⇒ { ((zeros at x=−2; x=3)),((no vertical asymptote)) :} defined for x∈R range: transforming y=((x^2 −x−6)/(x^2 −3x+4)) to x^2 −((3y−1)/(y−1))x+((2(2y+3))/(y−1))=0 D=−((7y^2 +14y−25)/(4(y−1)^2 ))≥0 for −1−((4(√(14)))/7)≤y≤−1+((4(√(14)))/7) ⇒ ⇒ { ((range=[−1−((4(√(14)))/7); −1+((4(√(14)))/7)])),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=5 ⇒ y=1 y′=−((2(x^2 −10x+11))/((x^2 −3x+4)^2 ))=0 at x=5±(√(14)) y^(′′) =((4(x^3 −15x^2 +33x−13))/((x^2 −3x+4)^3 )) y′′ { ((>0 at x=5−(√(14)) ⇒ local minimum)),((<0 at x=5+(√(14)) ⇒ local maximum)),((=0 at { ((x≈.506699)),((x≈2.06421)),((x≈12.4291)) :} ⇒ 3 turning points )) :} (3) y=((x^2 −x+6)/(x^2 −3x−4))= =((x^2 −x+6)/((x−4)(x+1))) ⇒ ⇒ { ((no real zeros)),((vertical asymptotes at x=−1; x=4)) :} defined for x∈R\{−1; 4} range: transforming y=((x^2 −x+6)/(x^2 −3x−4)) to x^2 −((3y−1)/(y−1))x−((2(2y+3))/(y−1))=0 D=((25y^2 +2y−23)/(4(y−1)^2 ))<0 for −1<y<((23)/(25)) ⇒ ⇒ { ((range=R\]−1; ((23)/(25))[)),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=−5 ⇒ y=1 y′=−((2(x^2 +10x−11))/((x−4)^2 (x+1)^2 ))=0 at x=−11; x=1 y′′=((4(x^3 +15x^2 −33x+53))/((x−4)^3 (x+1)^3 )) y′′ { ((>0 at x=−11 ⇒ local minimum)),((<0 at x=1 ⇒ local maximum)),((=0 at x≈−17.1098 ⇒ turning point)) :}

$$\mathrm{to}\:\mathrm{Sir}\:\mathrm{Jagoll}\:\left({and}\:{of}\:{course}\:{everybody}\:{else}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}= \\ $$$$=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{zeros}\:\mathrm{at}\:{x}=−\mathrm{2};\:{x}=\mathrm{3}}\\{\mathrm{vertical}\:\mathrm{asymptotes}\:\mathrm{at}\:{x}=−\mathrm{1};\:{x}=\mathrm{4}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1};\:\mathrm{4}\right\} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}−\frac{\mathrm{2}\left(\mathrm{2}{y}−\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{25}{y}^{\mathrm{2}} −\mathrm{46}{y}+\mathrm{25}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }>\mathrm{0}\forall{x}\in\mathbb{R}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{range}=\mathbb{R}}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{1}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{7}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{local}\:\mathrm{extremes} \\ $$$${y}''=\frac{\mathrm{4}\left({x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{21}{x}−\mathrm{25}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{0}\:\mathrm{at}\:{x}\approx\mathrm{1}.\mathrm{33131}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{turning}\:\mathrm{point} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}}= \\ $$$$=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}};\:{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}=\mathrm{0}\:\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{zeros}\:\mathrm{at}\:{x}=−\mathrm{2};\:{x}=\mathrm{3}}\\{\mathrm{no}\:\mathrm{vertical}\:\mathrm{asymptote}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}+\frac{\mathrm{2}\left(\mathrm{2}{y}+\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=−\frac{\mathrm{7}{y}^{\mathrm{2}} +\mathrm{14}{y}−\mathrm{25}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }\geqslant\mathrm{0}\:\mathrm{for}\:−\mathrm{1}−\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\leqslant{y}\leqslant−\mathrm{1}+\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{range}=\left[−\mathrm{1}−\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}};\:−\mathrm{1}+\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\right]}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{5}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{11}\right)}{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{2}} }=\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}\pm\sqrt{\mathrm{14}} \\ $$$${y}^{''} =\frac{\mathrm{4}\left({x}^{\mathrm{3}} −\mathrm{15}{x}^{\mathrm{2}} +\mathrm{33}{x}−\mathrm{13}\right)}{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{3}} } \\ $$$${y}''\begin{cases}{>\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}−\sqrt{\mathrm{14}}\:\Rightarrow\:\mathrm{local}\:\mathrm{minimum}}\\{<\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}+\sqrt{\mathrm{14}}\:\Rightarrow\:\mathrm{local}\:\mathrm{maximum}}\\{=\mathrm{0}\:\mathrm{at}\:\begin{cases}{{x}\approx.\mathrm{506699}}\\{{x}\approx\mathrm{2}.\mathrm{06421}}\\{{x}\approx\mathrm{12}.\mathrm{4291}}\end{cases}\:\Rightarrow\:\mathrm{3}\:\mathrm{turning}\:\mathrm{points}\:}\end{cases} \\ $$$$ \\ $$$$\left(\mathrm{3}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}= \\ $$$$=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}}\\{\mathrm{vertical}\:\mathrm{asymptotes}\:\mathrm{at}\:{x}=−\mathrm{1};\:{x}=\mathrm{4}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1};\:\mathrm{4}\right\} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}−\frac{\mathrm{2}\left(\mathrm{2}{y}+\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{25}{y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{23}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }<\mathrm{0}\:\mathrm{for}\:−\mathrm{1}<{y}<\frac{\mathrm{23}}{\mathrm{25}}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\left.\mathrm{range}=\mathbb{R}\backslash\right]−\mathrm{1};\:\frac{\mathrm{23}}{\mathrm{25}}\left[\right.}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=−\mathrm{5}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{11}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0}\:\mathrm{at}\:{x}=−\mathrm{11};\:{x}=\mathrm{1} \\ $$$${y}''=\frac{\mathrm{4}\left({x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} −\mathrm{33}{x}+\mathrm{53}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${y}''\begin{cases}{>\mathrm{0}\:\mathrm{at}\:{x}=−\mathrm{11}\:\Rightarrow\:\mathrm{local}\:\mathrm{minimum}}\\{<\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{1}\:\Rightarrow\:\mathrm{local}\:\mathrm{maximum}}\\{=\mathrm{0}\:\mathrm{at}\:{x}\approx−\mathrm{17}.\mathrm{1098}\:\Rightarrow\:\mathrm{turning}\:\mathrm{point}}\end{cases} \\ $$

Question Number 79879    Answers: 1   Comments: 5

Question Number 79876    Answers: 1   Comments: 1

f is derivable in R. 1) Demonstrate that if f is pair , f ′ is odd(unpair). 1) Demonstrate that if f is unpair , f ′ is pair. Please help me sirs

$${f}\:{is}\:{derivable}\:{in}\:\mathbb{R}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Demonstrate}\:\mathrm{that}\:\mathrm{if}\:{f}\:\mathrm{is}\:\mathrm{pair}\:,\:{f}\:'\:\mathrm{is}\:\mathrm{odd}\left(\mathrm{unpair}\right). \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Demonstrate}\:\mathrm{that}\:\mathrm{if}\:{f}\:{is}\:\mathrm{unpair}\:,\:{f}\:'\:\mathrm{is}\:\mathrm{pair}. \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{sirs} \\ $$

Question Number 79964    Answers: 0   Comments: 3

∫(tan^2 x+tan^4 x)dx=[tan^2 x=t]=∫(t^2 +t^4 )dx=(t^3 /3)+(t^5 /5)+c

$$\int\left(\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{tan}\:^{\mathrm{4}} {x}\right){dx}=\left[\mathrm{tan}\:^{\mathrm{2}} {x}={t}\right]=\int\left({t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right){dx}=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{{t}^{\mathrm{5}} }{\mathrm{5}}+{c} \\ $$

  Pg 1266      Pg 1267      Pg 1268      Pg 1269      Pg 1270      Pg 1271      Pg 1272      Pg 1273      Pg 1274      Pg 1275   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com