Question Number 226292 by Spillover last updated on 25/Nov/25
Answered by Frix last updated on 25/Nov/25
![=∫((cos x sin x)/(cos x +sin x))dx =^([t=tan (x/2)]) =4∫((t(t^2 −1))/((t^2 +1)^2 (t^2 −2t−1)))dt =^([decompose etc.]) =((t−1)/(t^2 +1))+((√2)/4)ln ∣((t−1−(√2))/(t−1+(√2)))∣ ...](https://www.tinkutara.com/question/Q226347.png)
$$=\int\frac{\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{dx}\:\overset{\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right]} {=} \\ $$$$=\mathrm{4}\int\frac{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)}{dt}\:\overset{\left[\mathrm{decompose}\:\mathrm{etc}.\right]} {=} \\ $$$$=\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\mid\frac{{t}−\mathrm{1}−\sqrt{\mathrm{2}}}{{t}−\mathrm{1}+\sqrt{\mathrm{2}}}\mid \\ $$$$… \\ $$
Answered by Ghisom_ last updated on 26/Nov/25
![∫(dx/(sec x +cosec x))=∫((cos x sin x)/(cos x +sin x))dx= [t=x+(π/4)] =((√2)/2)∫sin t dt−((√2)/4)∫(dt/(sin t))= =−((√2)/2)cos t −((√2)/4)ln tan (t/2) = =((sin x −cos x)/2)−((√2)/4)ln ∣(((√2)−cos x +sin x)/(cos x +sin x))∣ +C](https://www.tinkutara.com/question/Q226348.png)
$$\int\frac{{dx}}{\mathrm{sec}\:{x}\:+\mathrm{cosec}\:{x}}=\int\frac{\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\frac{\pi}{\mathrm{4}}\right] \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\mathrm{sin}\:{t}\:{dt}−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int\frac{{dt}}{\mathrm{sin}\:{t}}= \\ $$$$=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{cos}\:{t}\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:= \\ $$$$=\frac{\mathrm{sin}\:{x}\:−\mathrm{cos}\:{x}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\mid\frac{\sqrt{\mathrm{2}}−\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}\mid\:+{C} \\ $$