Question Number 96763 by bobhans last updated on 04/Jun/20
$$\int\:\frac{\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:\mathrm{dx}}{\mathrm{cos}\:\mathrm{x}}\:=\:? \\ $$
Answered by Sourav mridha last updated on 04/Jun/20
![=∫((1−cosx)/(2cos((x/2))cosx))dx =∫((sec^2 (x/2))/( (√(1−tan^2 (x/2)))))d((x/2))−∫sec(x/2)d((x/2)) =sin^(−1) [tan(x/2)]−ln[sec(x/2)+tan(x/2)]+c =](https://www.tinkutara.com/question/Q96768.png)
$$=\int\frac{\mathrm{1}−\boldsymbol{{cosx}}}{\mathrm{2}\boldsymbol{{cos}}\left(\frac{\boldsymbol{{x}}}{\mathrm{2}}\right)\boldsymbol{{cosx}}}\boldsymbol{{dx}} \\ $$$$=\int\frac{\boldsymbol{{sec}}^{\mathrm{2}} \frac{\boldsymbol{{x}}}{\mathrm{2}}}{\:\sqrt{\mathrm{1}−\boldsymbol{{tan}}^{\mathrm{2}} \frac{\boldsymbol{{x}}}{\mathrm{2}}}}\boldsymbol{{d}}\left(\frac{\boldsymbol{{x}}}{\mathrm{2}}\right)−\int\boldsymbol{{sec}}\frac{\boldsymbol{{x}}}{\mathrm{2}}\boldsymbol{{d}}\left(\frac{\boldsymbol{{x}}}{\mathrm{2}}\right) \\ $$$$=\mathrm{sin}^{−\mathrm{1}} \left[\boldsymbol{{tan}}\frac{\boldsymbol{{x}}}{\mathrm{2}}\right]−\boldsymbol{{ln}}\left[\boldsymbol{{sec}}\frac{\boldsymbol{{x}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{x}}}{\mathrm{2}}\right]+\boldsymbol{{c}} \\ $$$$= \\ $$
Commented by bobhans last updated on 05/Jun/20
![thank]you](https://www.tinkutara.com/question/Q96832.png)
$$\left.\mathrm{thank}\right]\mathrm{you} \\ $$
Answered by bobhans last updated on 05/Jun/20
![set x = 2t ⇒ dx = 2 dt K= ∫ ((2sin t tan t dt )/(cos 2t)) = ∫ ((2sin^2 t dt)/(cos t (1−2sin^2 t))) let q = sin t ⇒ dq = cos t dt K=∫ ((2q^2 dq)/((1−q^2 )(1−2q^2 ))) = 2∫ ((1/(1−2q^2 )) − (1/(1−q^2 ))) dq K= 2 [ (1/(2(√2))) ln ∣((1+q(√2))/(1−q(√2)))∣−(1/2)ln∣((1+q)/(1−q))∣ ]+ c K = (1/( (√2))) ln∣((1+(√2) sin ((x/2)))/(1−(√2) sin ((x/2))))∣ − ln ∣((1+sin ((x/2)))/(1−sin ((x/2))))∣ + c](https://www.tinkutara.com/question/Q96824.png)
$$\mathrm{set}\:\mathrm{x}\:=\:\mathrm{2t}\:\Rightarrow\:\mathrm{dx}\:=\:\mathrm{2}\:\mathrm{dt} \\ $$$$\mathrm{K}=\:\int\:\frac{\mathrm{2sin}\:\mathrm{t}\:\mathrm{tan}\:\mathrm{t}\:\mathrm{dt}\:}{\mathrm{cos}\:\mathrm{2t}}\:=\:\int\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \mathrm{t}\:\mathrm{dt}}{\mathrm{cos}\:\mathrm{t}\:\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{t}\right)} \\ $$$$\mathrm{let}\:\mathrm{q}\:=\:\mathrm{sin}\:\mathrm{t}\:\Rightarrow\:\mathrm{dq}\:=\:\mathrm{cos}\:\mathrm{t}\:\mathrm{dt} \\ $$$$\mathrm{K}=\int\:\frac{\mathrm{2q}^{\mathrm{2}} \:\mathrm{dq}}{\left(\mathrm{1}−\mathrm{q}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{2q}^{\mathrm{2}} \right)}\:=\:\mathrm{2}\int\:\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2q}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{q}^{\mathrm{2}} }\right)\:\mathrm{dq} \\ $$$$\mathrm{K}=\:\mathrm{2}\:\left[\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\mathrm{ln}\:\mid\frac{\mathrm{1}+\mathrm{q}\sqrt{\mathrm{2}}}{\mathrm{1}−\mathrm{q}\sqrt{\mathrm{2}}}\mid−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{\mathrm{1}+\mathrm{q}}{\mathrm{1}−\mathrm{q}}\mid\:\right]+\:\mathrm{c}\: \\ $$$$\mathrm{K}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{ln}\mid\frac{\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)}\mid\:−\:\mathrm{ln}\:\mid\frac{\mathrm{1}+\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)}\mid\:+\:\mathrm{c}\: \\ $$