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Question Number 25335 by mubeen897@hotmail.com last updated on 08/Dec/17

∫((tan(x))/((1+Cos(x))))dx=? Need help??

$$\int\frac{{tan}\left({x}\right)}{\left(\mathrm{1}+{Cos}\left({x}\right)\right)}{dx}=? \\ $$$${Need}\:{help}?? \\ $$$$ \\ $$

Answered by ajfour last updated on 08/Dec/17

I=∫((sin x)/(cos x(1+cos x)))dx let cos x=t ⇒ −sin xdx=dt ⇒ I=−∫(dt/(t(1+t))) =∫(dt/(1+t))−∫(dt/t) =ln ∣((1+t)/t)∣+c ⇒ I =ln ∣((1+cos x)/(cos x))∣+c .

$${I}=\int\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}\left(\mathrm{1}+\mathrm{cos}\:{x}\right)}{dx} \\ $$$${let}\:\:\mathrm{cos}\:{x}={t}\:\:\:\Rightarrow\:\:−\mathrm{sin}\:{xdx}={dt} \\ $$$$\Rightarrow\:{I}=−\int\frac{{dt}}{{t}\left(\mathrm{1}+{t}\right)}\:=\int\frac{{dt}}{\mathrm{1}+{t}}−\int\frac{{dt}}{{t}} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{ln}\:\mid\frac{\mathrm{1}+{t}}{{t}}\mid+{c} \\ $$$$\Rightarrow\:\:{I}\:=\mathrm{ln}\:\mid\frac{\mathrm{1}+\mathrm{cos}\:{x}}{\mathrm{cos}\:{x}}\mid+{c}\:. \\ $$

Answered by $@ty@m last updated on 08/Dec/17

=∫((sin x)/(cos x(1+cos x)))dx =∫((−dt)/(t(1+t))) =∫((−dt)/(t^2 +t+(1/4)−(1/4))) =∫((−dt)/((t+(1/2))^2 −((1/2))^2 )) =∫(dt/(((1/2))^2 −(t+(1/2))^2 )) =(1/(2×(1/2)))ln ∣(((1/2)+t+(1/2))/((1/2)−t−(1/2)))∣ =ln ∣((t+1)/t)∣+C

$$=\int\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}\left(\mathrm{1}+\mathrm{cos}\:{x}\right)}{dx} \\ $$$$=\int\frac{−{dt}}{{t}\left(\mathrm{1}+{t}\right)} \\ $$$$=\int\frac{−{dt}}{{t}^{\mathrm{2}} +{t}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$=\int\frac{−{dt}}{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\int\frac{{dt}}{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}}\mathrm{ln}\:\mid\frac{\frac{\mathrm{1}}{\mathrm{2}}+{t}+\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}−{t}−\frac{\mathrm{1}}{\mathrm{2}}}\mid \\ $$$$=\mathrm{ln}\:\mid\frac{{t}+\mathrm{1}}{{t}}\mid+{C} \\ $$

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