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1-sin-x-dx-




Question Number 4397 by moussapk last updated on 20/Jan/16
∫((1/(sin(x)))dx
$$\int\left(\frac{\mathrm{1}}{{sin}\left({x}\right)}{dx}\right. \\ $$
Answered by Yozzii last updated on 20/Jan/16
I=∫(1/(sinx))dx=∫cosecxdx  I=∫((cosecx(cosecx+cotx))/(cosecx+cotx))dx  I=∫((cosec^2 x+cotxcosecx)/(cosecx+cotx))dx  Let u=cotx+cosecx.  ∴ du=(−cosec^2 x−cosecxcotx)dx  −du=(cosec^2 x+cosecxcotx)dx  ∴I=−∫(1/u)du  I=−ln∣u∣+C    =−ln∣cosecx+cotx∣+C    =−ln∣(1/(sinx))+((cosx)/(sinx))∣+C    =−ln∣((1+cosx)/(sinx))∣+C    =−ln∣((2cos^2 0.5x)/(2sin0.5xcos0.5x))∣+C    =−ln∣cot0.5x∣+C  I=ln∣tan(1/2)x∣+C
$${I}=\int\frac{\mathrm{1}}{{sinx}}{dx}=\int{cosecxdx} \\ $$$${I}=\int\frac{{cosecx}\left({cosecx}+{cotx}\right)}{{cosecx}+{cotx}}{dx} \\ $$$${I}=\int\frac{{cosec}^{\mathrm{2}} {x}+{cotxcosecx}}{{cosecx}+{cotx}}{dx} \\ $$$${Let}\:{u}={cotx}+{cosecx}. \\ $$$$\therefore\:{du}=\left(−{cosec}^{\mathrm{2}} {x}−{cosecxcotx}\right){dx} \\ $$$$−{du}=\left({cosec}^{\mathrm{2}} {x}+{cosecxcotx}\right){dx} \\ $$$$\therefore{I}=−\int\frac{\mathrm{1}}{{u}}{du} \\ $$$${I}=−{ln}\mid{u}\mid+{C} \\ $$$$\:\:=−{ln}\mid{cosecx}+{cotx}\mid+{C} \\ $$$$\:\:=−{ln}\mid\frac{\mathrm{1}}{{sinx}}+\frac{{cosx}}{{sinx}}\mid+{C} \\ $$$$\:\:=−{ln}\mid\frac{\mathrm{1}+{cosx}}{{sinx}}\mid+{C} \\ $$$$\:\:=−{ln}\mid\frac{\mathrm{2}{cos}^{\mathrm{2}} \mathrm{0}.\mathrm{5}{x}}{\mathrm{2}{sin}\mathrm{0}.\mathrm{5}{xcos}\mathrm{0}.\mathrm{5}{x}}\mid+{C} \\ $$$$\:\:=−{ln}\mid{cot}\mathrm{0}.\mathrm{5}{x}\mid+{C} \\ $$$${I}={ln}\mid{tan}\frac{\mathrm{1}}{\mathrm{2}}{x}\mid+{C} \\ $$

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