Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 85667 by john santu last updated on 23/Mar/20

∫ (dx/(√(1−sin 2x)))

$$\int\:\frac{{dx}}{\sqrt{\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}}}\: \\ $$

Answered by som(math1967) last updated on 24/Mar/20

∫(dx/((cosx−sinx)))  ∫(dx/((√2)cos(x+(π/4))))  (1/(√2))∫sec(x+(π/4))dx  (1/(√2))ln∣tan((π/4) +(x/2)+(π/8))∣+C  (1/(√2))ln∣tan(((3π)/8) +(x/2))∣ +C

$$\int\frac{{dx}}{\left({cosx}−{sinx}\right)} \\ $$$$\int\frac{{dx}}{\sqrt{\mathrm{2}}{cos}\left({x}+\frac{\pi}{\mathrm{4}}\right)} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\int{sec}\left({x}+\frac{\pi}{\mathrm{4}}\right){dx} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{ln}\mid{tan}\left(\frac{\pi}{\mathrm{4}}\:+\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)\mid+{C} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{ln}\mid{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\:+\frac{{x}}{\mathrm{2}}\right)\mid\:+{C} \\ $$

Commented by jagoll last updated on 24/Mar/20

sir (√(1−sin 2x ))= ∣cos x−sin x∣  why (√(1−sin 2x))  = cos x−sin x ?

$$\mathrm{sir}\:\sqrt{\mathrm{1}−\mathrm{sin}\:\mathrm{2x}\:}=\:\mid\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\mid \\ $$$$\mathrm{why}\:\sqrt{\mathrm{1}−\mathrm{sin}\:\mathrm{2x}}\:\:=\:\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\:? \\ $$

Commented by jagoll last updated on 24/Mar/20

then ∫ sec x dx = ln ∣sec x+tan x∣ sir

$$\mathrm{then}\:\int\:\mathrm{sec}\:\mathrm{x}\:\mathrm{dx}\:=\:\mathrm{ln}\:\mid\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}\mid\:\mathrm{sir} \\ $$

Commented by som(math1967) last updated on 24/Mar/20

secx+tanx=((1+sinx)/(cosx))=((sin(x/2)+cos(x/2))/(cos(x/2)−sin(x/2)))    =((tan(x/2)+1)/(1−tan(x/2)))=tan((x/2) +(π/4))

$${secx}+{tanx}=\frac{\mathrm{1}+{sinx}}{{cosx}}=\frac{{sin}\frac{{x}}{\mathrm{2}}+{cos}\frac{{x}}{\mathrm{2}}}{{cos}\frac{{x}}{\mathrm{2}}−{sin}\frac{{x}}{\mathrm{2}}} \\ $$$$\:\:=\frac{{tan}\frac{{x}}{\mathrm{2}}+\mathrm{1}}{\mathrm{1}−{tan}\frac{{x}}{\mathrm{2}}}={tan}\left(\frac{{x}}{\mathrm{2}}\:+\frac{\pi}{\mathrm{4}}\right) \\ $$

Commented by jagoll last updated on 24/Mar/20

oo same sir. but i don′t agree  with (√(1−sin 2x)) = cos x−sin x sir

$$\mathrm{oo}\:\mathrm{same}\:\mathrm{sir}.\:\mathrm{but}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{agree} \\ $$$$\mathrm{with}\:\sqrt{\mathrm{1}−\mathrm{sin}\:\mathrm{2x}}\:=\:\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\:\mathrm{sir} \\ $$

Commented by jagoll last updated on 24/Mar/20

i got two solution

$$\mathrm{i}\:\mathrm{got}\:\mathrm{two}\:\mathrm{solution}\: \\ $$

Commented by som(math1967) last updated on 24/Mar/20

yes two solution possible  I use only (cosx−sinx)

$${yes}\:{two}\:{solution}\:{possible} \\ $$$${I}\:{use}\:{only}\:\left({cosx}−{sinx}\right) \\ $$

Commented by jagoll last updated on 24/Mar/20

yes sir. deal

$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{deal} \\ $$

Commented by john santu last updated on 24/Mar/20

yes have 2 solution

$${yes}\:{have}\:\mathrm{2}\:{solution} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com