Integration Questions

Question Number 118891 by cantor last updated on 20/Oct/20

$$\:\int_{\mathrm{0}} ^{\pi} \boldsymbol{{arctan}}\left(\mathrm{3}^{\boldsymbol{{cosx}}} \right)\boldsymbol{{dx}}=??? \\$$$$\\$$$$\boldsymbol{{please}}\:\boldsymbol{{help}} \\$$

Answered by Dwaipayan Shikari last updated on 20/Oct/20

$$\int_{\mathrm{0}} ^{\pi} {tan}^{−\mathrm{1}} \left(\mathrm{3}^{{cosx}} \right){dx}=\int_{\mathrm{0}} ^{\pi} {tan}^{−\mathrm{1}} \left(\mathrm{3}^{−{cosx}} \right){dx}={I}\:\:\:\:\:\:\left({Using}\:\int_{\mathrm{0}} ^{\pi} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\pi} {f}\left(\pi−{x}\right){dx}\right. \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi} {tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}^{{cosx}} +\mathrm{3}^{−{cosx}} }{\mathrm{1}−\mathrm{3}^{{cosx}−{cosx}} }\right){dx} \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi} {tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}^{{cosx}} +\mathrm{3}^{−{cosx}} }{{z}}\right){dx}\:\:\:\:\:\:\:\:\:\:\:\:{z}\rightarrow\mathrm{0}\:\:{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}^{{cosx}} +\mathrm{3}^{−{cosx}} }{{z}}\right)=\frac{\pi}{\mathrm{2}} \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi} \frac{\pi}{\mathrm{2}}{dx}\:\:\:\:\:\:\:\:\: \\$$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\$$$$\\$$

Commented by mnjuly1970 last updated on 20/Oct/20

$${let}\:{me}\:{explain}. \\$$$${note}::\:{tan}^{−\mathrm{1}} \left({x}\right)+{tan}^{−\mathrm{1}} \left({y}\right)={tan}^{−\mathrm{1}} \left(\frac{{x}+{y}}{\mathrm{1}−{xy}}\right) \\$$$${I}=\int_{\mathrm{0}\:} ^{\:\pi} {tan}^{−\mathrm{1}} \left(\mathrm{3}^{{cos}\left({x}\right)} \right){dx}.....\left(\mathrm{1}\right) \\$$$${I}\:\:\:\overset{\left[\int_{{a}} ^{\:{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{\:{b}} {f}\left({a}+{b}−{x}\right){dx}\right]} {=}\int_{\mathrm{0}} ^{\:\pi} {tan}^{−\mathrm{1}} \left(\mathrm{3}^{−{cos}\left({x}\right)} \right)...\left(\mathrm{2}\right) \\$$$$\therefore\left(\mathrm{1}\right)+\left(\mathrm{2}\right)::\: \\$$$$\mathrm{2}{I}\overset{{note}} {=}\int_{\mathrm{0}} ^{\:\:\pi} {tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}^{{cos}\left({x}\right)} +\mathrm{3}^{−{cos}\left({x}\right)} }{\left[\mathrm{1}−\mathrm{3}^{{cos}\left({x}\right)} .\mathrm{3}^{−{cos}\left({x}\right)} \right]=\mathrm{0}}\right) \\$$$$\:\:\:\:\:\:\: \\$$$$=\int_{\mathrm{0}} ^{\:\pi} \left[{tan}^{−\mathrm{1}} \left(\infty\right)=\frac{\pi}{\mathrm{2}}\right]{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\$$$$\therefore\:\:{I}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\:\:\:\checkmark\checkmark \\$$$$\:\:\:\:\:\:\:\:\:\:\:...\:{m}.{n}.\mathrm{1970}... \\$$

Commented by cantor last updated on 20/Oct/20

$$\boldsymbol{{please}}\:\boldsymbol{{i}}\:\boldsymbol{{don}}'\boldsymbol{{t}}\:\boldsymbol{{understand}} \\$$$$\boldsymbol{{the}}\:\mathrm{1}^{\boldsymbol{{st}}} \:\boldsymbol{{line}}\:\boldsymbol{{please}}\:\boldsymbol{{explaint}} \\$$

Commented by benjo_mathlover last updated on 20/Oct/20

$${replace}\:{x}\:{by}\:\pi−{x}\: \\$$

Answered by mindispower last updated on 20/Oct/20

$$=\int_{\mathrm{0}} ^{\pi} {arctan}\left(\mathrm{3}^{{cos}\left(\pi−{x}\right)} \right){dx}=\int{arctan}\left(\frac{\mathrm{1}}{\mathrm{3}^{{cos}\left({x}\right)} }\right) \\$$$${arctan}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\pi}{\mathrm{2}}−{arctan}\left({x}\right),{x}>\mathrm{0} \\$$$$..... \\$$