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Question Number 201224 by Calculusboy last updated on 02/Dec/23

∫ (1/( (√(1+sinx))))dx

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

Answered by witcher3 last updated on 02/Dec/23

(√(1+sin(x)))=hint :∣sin((x/2))+cos((x/2))∣

$$\sqrt{\mathrm{1}+\mathrm{sin}\left(\mathrm{x}\right)}=\mathrm{hint}\::\mid\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mid \\ $$

Answered by Sutrisno last updated on 04/Dec/23

=∫(1/( (√(sin^2 (1/2)x+cos^2 (1/2)x+2sin(1/2)xcos(1/2)x))))dx  =∫(1/( (√((sin(1/2)x+cos(1/2)x)^2 ))))dx  =∫(1/( sin(1/2)x+cos(1/2)x))dx  =∫(1/( (√2)cos((1/2)x−(π/4))))dx  =∫((sec((1/2)x−(π/4)))/( (√2)))dx  =(2/( (√2)))ln∣sec((1/2)x−(π/4))+tan((1/2)x−(π/4))∣+c

$$=\int\frac{\mathrm{1}}{\:\sqrt{{sin}^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{x}+{cos}^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{2}{sin}\frac{\mathrm{1}}{\mathrm{2}}{xcos}\frac{\mathrm{1}}{\mathrm{2}}{x}}}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{\left({sin}\frac{\mathrm{1}}{\mathrm{2}}{x}+{cos}\frac{\mathrm{1}}{\mathrm{2}}{x}\right)^{\mathrm{2}} }}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\:{sin}\frac{\mathrm{1}}{\mathrm{2}}{x}+{cos}\frac{\mathrm{1}}{\mathrm{2}}{x}}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\pi}{\mathrm{4}}\right)}{dx} \\ $$$$=\int\frac{{sec}\left(\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\pi}{\mathrm{4}}\right)}{\:\sqrt{\mathrm{2}}}{dx} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}{ln}\mid{sec}\left(\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\pi}{\mathrm{4}}\right)+{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\pi}{\mathrm{4}}\right)\mid+{c} \\ $$

Commented by Calculusboy last updated on 04/Dec/23

thanks sir

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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