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Question Number 79520 by TawaTawa last updated on 25/Jan/20

Commented by mathmax by abdo last updated on 26/Jan/20

$$\mathrm{\Omega }={\int }_0^{\frac{2}{5}}\frac{{cos}^{2}x}{{cos}^2(x-\frac{1}{5})}dx={\int }_0^{\frac{2}{5}}\frac{1+cos\left(2x\right)}{1+cos(2x-\frac{2}{5})}dx$$$${=}_{2x=t}\frac{1}{2}{\int }_0^{\frac{4}{5}}\frac{1+cost}{1+cos(t-\frac{2}{5})}dt{=}_{t-\frac{2}{5}=u}\frac{1}{2}{\int }_{-\frac{2}{5}}^{\frac{2}{5}}\frac{1+cos(u+\frac{2}{5})}{1+cosu}du$$$$\Rightarrow 2\mathrm{\Omega }={\int }_{-\frac{2}{5}}^{\frac{2}{5}}\frac{1+cos\left(\frac{2}{5}\right)cosu-sin\left(\frac{2}{5}\right)sinu}{1+cosu}du$$$$={\int }_{-\frac{2}{5}}^{\frac{2}{5}}\frac{du}{1+cosu}+cos\left(\frac{2}{5}\right){\int }_{-\frac{2}{5}}^{\frac{2}{5}}\frac{cosu}{1+cosu}du-sin\left(\frac{2}{5}\right){\int }_{-\frac{2}{5}}^{\frac{2}{5}}\frac{sinu}{1+codu}du(\to 0)$$$$\mathrm{\Omega }={\int }_0^{\frac{2}{5}}\frac{du}{1+cosu}+cos\left(\frac{2}{5}\right){\int }_0^{\frac{2}{5}}\frac{cosu}{1+cosu}du$$$${\int }_0^{\frac{2}{5}}\frac{du}{1+cosu}du{=}_{tan\left(\frac{u}{2}\right)=z}{\int }_0^{tan\left(\frac{1}{5}\right)}\frac{1}{1+\frac{1-z^2}{1+z^2}}\times \frac{2dz}{1+z^2}$$$$=2{\int }_0^{tan\left(\frac{1}{5}\right)}\frac{dz}{1+z^2+1-z^2}=tan\left(\frac{1}{5}\right)$$$${\int }_0^{\frac{2}{5}}\frac{cosu}{1+cosu}du={\int }_0^{\frac{2}{5}}\frac{1+cosu-1}{1+cosu}du=\frac{2}{5}-tan\left(\frac{1}{5}\right)\Rightarrow$$$$\mathrm{\Omega }=tan\left(\frac{1}{5}\right)+cos\left(\frac{2}{5}\right)(\frac{2}{5}-tan\left(\frac{1}{5}\right))$$

Commented by TawaTawa last updated on 26/Jan/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by MJS last updated on 25/Jan/20

$$\int \frac{{\cos }^{2}x}{{\cos }^2(x-\frac{1}{5})}dx=$$$$[t=x-\frac{1}{5}\to dx=dt]$$$$\int \frac{{\cos }^2(t+\frac{1}{5})}{{\cos }^{2}t}dt=$$$$=\frac{1+\cos \frac{2}{5}}{2}\int dt-\sin \frac{2}{5}\int \tan tdt+\frac{1-\cos \frac{2}{5}}{2}\int {\tan }^{2}tdt=$$$$\mathrm{and}\mathrm{these}\mathrm{are}\mathrm{easy}\mathrm{to}\mathrm{solve}$$

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