dx-1-sin-2x-

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

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

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

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

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