Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 176570 by mathlove last updated on 21/Sep/22

(1)  ∫^(π/2) _(π/3) ((1+sinx)/(cosx)) dx=?

$$\left(\mathrm{1}\right)\:\:\underset{\frac{\pi}{\mathrm{3}}} {\int}^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{sinx}}{{cosx}}\:{dx}=? \\ $$

Answered by Peace last updated on 21/Sep/22

∫((1+sin(x))/(cos(x)))dx=∫((cos(x))/(cos^2 (x)))+∫((sin(x))/(cos(x)))dx  =∫((cos(x))/(1−sin^2 (x)))dx−∫((d(cos(x)))/(cos(x)))  =∫((d(sinx))/(1−sin^2 (x)))−ln∣cos(x)∣  =_ argth^− (sin(x))−ln∣cos(x)∣+c

$$\int\frac{\mathrm{1}+{sin}\left({x}\right)}{{cos}\left({x}\right)}{dx}=\int\frac{{cos}\left({x}\right)}{{cos}^{\mathrm{2}} \left({x}\right)}+\int\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)}{dx} \\ $$$$=\int\frac{{cos}\left({x}\right)}{\mathrm{1}−{sin}^{\mathrm{2}} \left({x}\right)}{dx}−\int\frac{{d}\left({cos}\left({x}\right)\right)}{{cos}\left({x}\right)} \\ $$$$=\int\frac{{d}\left({sinx}\right)}{\mathrm{1}−{sin}^{\mathrm{2}} \left({x}\right)}−{ln}\mid{cos}\left({x}\right)\mid \\ $$$$\underset{} {=}{argth}^{−} \left({sin}\left({x}\right)\right)−{ln}\mid{cos}\left({x}\right)\mid+{c} \\ $$

Answered by BaliramKumar last updated on 23/Sep/22

∫[(1/2)sec^2 ((x/2))+tan((x/2))]dx = [tan((x/2))−2ln∣cos((x/2))∣]_(π/3) ^(π/2)   = 1+ln(2)−(1/( (√3)))+2ln(((√3)/2))  = 1−ln(2)−(1/( (√3)))+ln(3)  = 1−(1/( (√3))) + ln((3/2))

$$\int\left[\frac{\mathrm{1}}{\mathrm{2}}{sec}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)+{tan}\left(\frac{{x}}{\mathrm{2}}\right)\right]{dx}\:=\:\left[{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}{ln}\mid{cos}\left(\frac{{x}}{\mathrm{2}}\right)\mid\underset{\frac{\pi}{\mathrm{3}}} {\overset{\frac{\pi}{\mathrm{2}}} {\right]}} \\ $$$$=\:\mathrm{1}+{ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\mathrm{2}{ln}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$=\:\mathrm{1}−{ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+{ln}\left(\mathrm{3}\right) \\ $$$$=\:\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\:{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com