Question Number 186910 by Spillover last updated on 11/Feb/23
$$ \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{{a}} \sqrt{\frac{\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:\mathrm{2}{a}}{\mathrm{cos}\:\mathrm{2}{x}+\mathrm{1}}}\:{dx}=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\:{a}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by witcher3 last updated on 16/Feb/23
![cos(2x)−cos(2a)=2cos^2 (x)−2cos^2 (a) cos(2x)+1=2cos^2 (x) ⇔(1/(cos(x)))(√(cos^2 (x)−cos^2 (a)))dx ∫_0 ^a ((cos(x)(√(sin^2 (a)−sin^2 (x))))/(1−sin^2 (x)))dx =∫_0 ^(sin(a)) (√(sin^2 (a)−t^2 )),(dt/(1−t^2 )) t=sin(a)sin(w) ∫_0 ^(π/2) ((sin^2 (a)cos^2 (w))/(1−sin^2 (a)sin^z (w)))dw =∫((sin^2 (a)−sin^2 (a)sin^2 (w))/(1−sin^2 (a)sin^2 (w)))dw =(π/2)−cos^2 (a)∫_0 ^(π/2) (dw/(1−sin^2 (a)sin^2 (w))) =(π/2)−cos^2 (a)∫(dw/(cos^2 (a)+sin^2 (a)cos^2 (w))) =(π/2)−∫((dtg(w))/(1+tg^2 (a)+tg^2 (w))) =(π/2)−(1/( (√(1+tg^2 (a))))).tan^(−1) (((tg(w))/( (√(1+tg^2 (a))))))]_0 ^(π/2) =(π/2)−cos(a)(π/2)=(π/2)(1−cos(a))](https://www.tinkutara.com/question/Q187360.png)
$$\mathrm{cos}\left(\mathrm{2x}\right)−\mathrm{cos}\left(\mathrm{2a}\right)=\mathrm{2cos}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{2cos}^{\mathrm{2}} \left(\mathrm{a}\right) \\ $$$$\mathrm{cos}\left(\mathrm{2x}\right)+\mathrm{1}=\mathrm{2cos}^{\mathrm{2}} \left(\mathrm{x}\right) \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{x}\right)}\sqrt{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{cos}^{\mathrm{2}} \left(\mathrm{a}\right)}\mathrm{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{a}} \frac{\mathrm{cos}\left(\mathrm{x}\right)\sqrt{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{a}\right)−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{sin}\left(\mathrm{a}\right)} \sqrt{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{a}\right)−\mathrm{t}^{\mathrm{2}} },\frac{\mathrm{dt}}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} } \\ $$$$\mathrm{t}=\mathrm{sin}\left(\mathrm{a}\right)\mathrm{sin}\left(\mathrm{w}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{a}\right)\mathrm{cos}^{\mathrm{2}} \left(\mathrm{w}\right)}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{a}\right)\mathrm{sin}^{\mathrm{z}} \left(\mathrm{w}\right)}\mathrm{dw} \\ $$$$=\int\frac{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{a}\right)−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{a}\right)\mathrm{sin}^{\mathrm{2}} \left(\mathrm{w}\right)}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{a}\right)\mathrm{sin}^{\mathrm{2}} \left(\mathrm{w}\right)}\mathrm{dw} \\ $$$$=\frac{\pi}{\mathrm{2}}−\mathrm{cos}^{\mathrm{2}} \left(\mathrm{a}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{dw}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{a}\right)\mathrm{sin}^{\mathrm{2}} \left(\mathrm{w}\right)} \\ $$$$=\frac{\pi}{\mathrm{2}}−\mathrm{cos}^{\mathrm{2}} \left(\mathrm{a}\right)\int\frac{\mathrm{dw}}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{a}\right)+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{a}\right)\mathrm{cos}^{\mathrm{2}} \left(\mathrm{w}\right)} \\ $$$$=\frac{\pi}{\mathrm{2}}−\int\frac{\mathrm{dtg}\left(\mathrm{w}\right)}{\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\mathrm{a}\right)+\mathrm{tg}^{\mathrm{2}} \left(\mathrm{w}\right)} \\ $$$$\left.=\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\mathrm{a}\right)}}.\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{tg}\left(\mathrm{w}\right)}{\:\sqrt{\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\mathrm{a}\right)}}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\pi}{\mathrm{2}}−\mathrm{cos}\left(\mathrm{a}\right)\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{a}\right)\right) \\ $$