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Question Number 149996 by tabata last updated on 08/Aug/21

(1) ∫ (dx/(1+tanx)) (2)∫ ((√(tanx))/(sinx cosx))dx

$$\left(\mathrm{1}\right)\:\int\:\:\frac{{dx}}{\mathrm{1}+{tanx}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\int\:\:\frac{\sqrt{{tanx}}}{{sinx}\:{cosx}}{dx} \\ $$

Answered by mindispower last updated on 08/Aug/21

=∫(1/(cos^2 (x))).((√(tg(x)))/(tg(x)))dx=∫(1/(cos^2 (x))).(1/( (√(tg(x)))))dx =2∫((dtg(x))/(2(√(tg(x)))))=2(√(tg(x)))+c

$$=\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}.\frac{\sqrt{{tg}\left({x}\right)}}{{tg}\left({x}\right)}{dx}=\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}.\frac{\mathrm{1}}{\:\sqrt{{tg}\left({x}\right)}}{dx} \\ $$$$=\mathrm{2}\int\frac{{dtg}\left({x}\right)}{\mathrm{2}\sqrt{{tg}\left({x}\right)}}=\mathrm{2}\sqrt{{tg}\left({x}\right)}+{c} \\ $$

Commented by tabata last updated on 08/Aug/21

thank you alot sir

$${thank}\:{you}\:{alot}\:{sir} \\ $$

Answered by mindispower last updated on 08/Aug/21

(1)=∫((cos(x))/(sin(x)+cos(x)))dx=(1/2)(∫((cos(x)+sin(x))/(sin(x)+cos(x)))dx+∫((cos(x)−sin(x))/(cos(x)+sin(x)))dx) =(x/2)+(1/2)ln∣cos(x)+sin(x)∣+c =(x/2)+((ln∣(√2)sin(x+(π/2))∣)/2)+c

$$\left(\mathrm{1}\right)=\int\frac{{cos}\left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left(\int\frac{{cos}\left({x}\right)+{sin}\left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}+\int\frac{{cos}\left({x}\right)−{sin}\left({x}\right)}{{cos}\left({x}\right)+{sin}\left({x}\right)}{dx}\right) \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{cos}\left({x}\right)+{sin}\left({x}\right)\mid+{c} \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{{ln}\mid\sqrt{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{2}}\right)\mid}{\mathrm{2}}+{c} \\ $$

Commented by puissant last updated on 08/Aug/21

nice sir mindispower...

$${nice}\:{sir}\:{mindispower}... \\ $$

Commented by tabata last updated on 08/Aug/21

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by puissant last updated on 08/Aug/21

2) Q=∫((√(tanx))/(sinx cosx))dx u=(√(tanx))→u^2 =tanx→2udu=(1/(cos^2 x))dx → dx=2ucos^2 xdu ⇒ Q=∫(u/(sinx cosx))2ucos^2 xdu ⇒ Q=2∫u^2 cotanxdu ⇒ Q=2∫u^2 (1/u^2 )du = 2∫du = 2u+C.. ∵ Q = 2(√(tanx))+C..

$$\left.\mathrm{2}\right) \\ $$$${Q}=\int\frac{\sqrt{{tanx}}}{{sinx}\:{cosx}}{dx} \\ $$$${u}=\sqrt{{tanx}}\rightarrow{u}^{\mathrm{2}} ={tanx}\rightarrow\mathrm{2}{udu}=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\rightarrow\:{dx}=\mathrm{2}{ucos}^{\mathrm{2}} {xdu} \\ $$$$\Rightarrow\:{Q}=\int\frac{{u}}{{sinx}\:{cosx}}\mathrm{2}{ucos}^{\mathrm{2}} {xdu} \\ $$$$\Rightarrow\:{Q}=\mathrm{2}\int{u}^{\mathrm{2}} {cotanxdu} \\ $$$$\Rightarrow\:{Q}=\mathrm{2}\int{u}^{\mathrm{2}} \frac{\mathrm{1}}{{u}^{\mathrm{2}} }{du}\:=\:\mathrm{2}\int{du}\:=\:\mathrm{2}{u}+{C}.. \\ $$$$\:\:\:\:\because\:{Q}\:=\:\mathrm{2}\sqrt{{tanx}}+{C}.. \\ $$$$ \\ $$

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