sinx-cosx-1-sin2x-dx-

Question Number 105969 by Ar Brandon last updated on 02/Aug/20

$$\int\frac{\mathrm{sinx}+\mathrm{cosx}}{\:\sqrt{\mathrm{1}+\mathrm{sin2x}}}\mathrm{dx} \\ $$

Answered by bobhans last updated on 02/Aug/20

∫ ((sin x+cos x)/( (√((sin x+cos x)^2 )))) dx = ∫(( (sin x+cos x)dx)/(∣sin x+cos x∣)) = ± ∫ dx = ±x + C

$$\int\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\:\sqrt{\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }}\:\mathrm{dx}\:=\:\int\frac{\:\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)\mathrm{dx}}{\mid\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\mid} \\ $$$$=\:\pm\:\int\:\mathrm{dx}\:=\:\pm\mathrm{x}\:+\:\mathrm{C} \\ $$

Commented by Ar Brandon last updated on 02/Aug/20

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{idea}. \\ $$

Commented by bobhans last updated on 02/Aug/20

$$\mathrm{sorry}\:\mathrm{typo} \\ $$

Commented by malwaan last updated on 02/Aug/20

$${sir}\:{bobhans} \\ $$$${please}\:{repost}\:{the}\:{answer} \\ $$$${thanks} \\ $$

Answered by Ar Brandon last updated on 02/Aug/20

I=∫((sinx+cosx)/( (√(1+sin2x))))dx=∫((sinx+cosx)/( (√(cos^2 x+2sinxcosx+sin^2 x))))dx =∫((sinx+cosx)/( (√((cosx+sinx)^2 ))))dx=∫((sinx+cosx)/(sinx+cosx))dx=∫dx =x+C

$$\mathcal{I}=\int\frac{\mathrm{sinx}+\mathrm{cosx}}{\:\sqrt{\mathrm{1}+\mathrm{sin2x}}}\mathrm{dx}=\int\frac{\mathrm{sinx}+\mathrm{cosx}}{\:\sqrt{\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{2sinxcosx}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx} \\ $$$$\:\:\:=\int\frac{\mathrm{sinx}+\mathrm{cosx}}{\:\sqrt{\left(\mathrm{cosx}+\mathrm{sinx}\right)^{\mathrm{2}} }}\mathrm{dx}=\int\frac{\mathrm{sinx}+\mathrm{cosx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}=\int\mathrm{dx} \\ $$$$\:\:\:=\mathrm{x}+\mathcal{C} \\ $$

Commented by prakash jain last updated on 02/Aug/20

Strictly speaking this is not correct solution for cases when (sin x+cos x)<0.

$$\mathrm{Strictly}\:\mathrm{speaking}\:\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{correct} \\ $$$$\mathrm{solution}\:\mathrm{for}\:\mathrm{cases}\:\mathrm{when}\:\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)<\mathrm{0}. \\ $$

Commented by Ar Brandon last updated on 02/Aug/20

OK Sir, thanks. I didn't take that into consideration. You're right.

Answered by bemath last updated on 02/Aug/20

∫ ((sin x+cos x)/(∣sin x+cos x∣)) dx = ∫± dx = ± x + C

$$\int\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mid\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\mid}\:{dx}\:=\:\int\pm\:{dx}\: \\ $$$$=\:\pm\:{x}\:+\:{C} \\ $$

Commented by prakash jain last updated on 02/Aug/20

$$\mathrm{or}\:{x}\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mid\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\mid}+{C} \\ $$

Commented by 1549442205PVT last updated on 02/Aug/20

I agree to Sir′s idea since ∫_a ^( b) ((sinx+cosx)/( (√(1+sin2x))))dx=( x((sin x+cos x)/(∣sin x+cos x∣))+C)∣_a ^b is true for (a,b)∈{(1,2),(3,5)}

$$\mathrm{I}\:\mathrm{agree}\:\mathrm{to}\:\mathrm{Sir}'\mathrm{s}\:\:\mathrm{idea}\:\mathrm{since}\: \\ $$$$\int_{\mathrm{a}} ^{\:\mathrm{b}} \frac{\mathrm{sinx}+\mathrm{cosx}}{\:\sqrt{\mathrm{1}+\mathrm{sin2x}}}\mathrm{dx}=\left(\:{x}\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mid\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\mid}+{C}\right)\mid_{\mathrm{a}} ^{\mathrm{b}} \\ $$$$\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\left(\mathrm{a},\mathrm{b}\right)\in\left\{\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{3},\mathrm{5}\right)\right\} \\ $$

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