Question-202212

Question Number 202212 by Calculusboy last updated on 22/Dec/23

Commented by BOYQOBILOV last updated on 23/Dec/23

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Answered by shunmisaki007 last updated on 23/Dec/23

I=∫_0 ^π (1/(1+(sin(x))^(tan(x)) ))dx =∫_0 ^(π/2) (1/(1+(sin(x))^(tan(x)) ))dx+∫_(π/2) ^π (1/(1+(sin(x))^(tan(x)) ))dx Consider ∫_(π/2) ^π (1/(1+(sin(x))^(tan(x)) ))dx. ⇒ Let u=π−x, du=−dx ⇒ ∫_(π/2) ^π (1/(1+(sin(x))^(tan(x)) ))dx =−∫_(π/2) ^0 (1/(1+(sin(π−u))^(tan(π−u)) ))du =∫_0 ^(π/2) (1/(1+(sin(u))^(−tan(u)) ))du ⇒I=∫_0 ^(π/2) (1/(1+(sin(x))^(tan(x)) ))dx+∫_0 ^(π/2) (1/(1+(sin(x))^(tan(x)) ))dx =∫_0 ^(π/2) (1/(1+(sin(x))^(tan(x)) ))+(1/(1+(sin(x))^(−tan(x)) ))dx =∫_0 ^(π/2) ((1+(sin(x))^(tan(x)) +1+(sin(x))^(tan(x)) )/(1+(sin(x))^(tan(x)) +(sin(x))^(−tan(x)) +1))dx =∫_0 ^(π/2) dx ∴ I=(π/2) ★

$$\:\:\:{I}=\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{sin}\left({x}\right)\right)^{\mathrm{tan}\left({x}\right)} }{dx} \\ $$$$\:\:\:\:\:\:=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{sin}\left({x}\right)\right)^{\mathrm{tan}\left({x}\right)} }{dx}+\underset{\frac{\pi}{\mathrm{2}}} {\overset{\pi} {\int}}\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{sin}\left({x}\right)\right)^{\mathrm{tan}\left({x}\right)} }{dx} \\ $$$$\mathrm{Consider}\:\underset{\frac{\pi}{\mathrm{2}}} {\overset{\pi} {\int}}\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{sin}\left({x}\right)\right)^{\mathrm{tan}\left({x}\right)} }{dx}. \\ $$$$\Rightarrow\:\mathrm{Let}\:{u}=\pi−{x},\:{du}=−{dx} \\ $$$$\Rightarrow\:\underset{\frac{\pi}{\mathrm{2}}} {\overset{\pi} {\int}}\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{sin}\left({x}\right)\right)^{\mathrm{tan}\left({x}\right)} }{dx} \\ $$$$\:\:\:\:\:\:=−\underset{\frac{\pi}{\mathrm{2}}} {\overset{\mathrm{0}} {\int}}\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{sin}\left(\pi−{u}\right)\right)^{\mathrm{tan}\left(\pi−{u}\right)} }{du}\: \\ $$$$\:\:\:\:\:\:=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{sin}\left({u}\right)\right)^{−\mathrm{tan}\left({u}\right)} }{du} \\ $$$$\Rightarrow{I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{sin}\left({x}\right)\right)^{\mathrm{tan}\left({x}\right)} }{dx}+\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{sin}\left({x}\right)\right)^{\mathrm{tan}\left({x}\right)} }{dx} \\ $$$$\:\:\:\:\:\:=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{sin}\left({x}\right)\right)^{\mathrm{tan}\left({x}\right)} }+\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{sin}\left({x}\right)\right)^{−\mathrm{tan}\left({x}\right)} }{dx} \\ $$$$\:\:\:\:\:\:=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{1}+\left(\mathrm{sin}\left({x}\right)\right)^{\mathrm{tan}\left({x}\right)} +\mathrm{1}+\left(\mathrm{sin}\left({x}\right)\right)^{\mathrm{tan}\left({x}\right)} }{\mathrm{1}+\left(\mathrm{sin}\left({x}\right)\right)^{\mathrm{tan}\left({x}\right)} +\left(\mathrm{sin}\left({x}\right)\right)^{−\mathrm{tan}\left({x}\right)} +\mathrm{1}}{dx} \\ $$$$\:\:\:\:\:\:=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{dx} \\ $$$$\therefore\:{I}=\frac{\pi}{\mathrm{2}}\:\:\:\bigstar \\ $$

Commented by Calculusboy last updated on 23/Dec/23

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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