Question Number 195602 by mokys last updated on 05/Aug/23
$${solve}\:\int\:\frac{{dx}}{{sin}^{\mathrm{10}} \left({x}\right)+{cos}^{\mathrm{10}} \left({x}\right)} \\ $$
Answered by Frix last updated on 06/Aug/23
![∫(dx/(sin^(10) x +cos^(10) x))= =64∫(dx/(5cos^2 4x +30cos 4x +29))= =(5/( (√5)))∫(dx/(3−(4/( (√5)))+cos 4x))−(8/( (√5)))∫(dx/(3+(4/( (√5)))+cos 4x))= [using t=tan 2x] =(4+2(√5))∫(dt/(t^2 +6+2(√5)))+(4−2(√5))∫(dt/(t^2 +6−2(√5)))= =−(((3+(√5))/2)tan^(−1) (((1−(√5))t)/4) +((3−(√5))/2)tan^(−1) (((1+(√5))t)/4))= =−(((3+(√5))/2)tan^(−1) (((1−(√5))tan 2x)/4) +((3−(√5))/2)tan^(−1) (((1+(√5))tan 2x)/4))+C](https://www.tinkutara.com/question/Q195614.png)
$$\int\frac{{dx}}{\mathrm{sin}^{\mathrm{10}} \:{x}\:+\mathrm{cos}^{\mathrm{10}} \:{x}}= \\ $$$$=\mathrm{64}\int\frac{{dx}}{\mathrm{5cos}^{\mathrm{2}} \:\mathrm{4}{x}\:+\mathrm{30cos}\:\mathrm{4}{x}\:+\mathrm{29}}= \\ $$$$=\frac{\mathrm{5}}{\:\sqrt{\mathrm{5}}}\int\frac{{dx}}{\mathrm{3}−\frac{\mathrm{4}}{\:\sqrt{\mathrm{5}}}+\mathrm{cos}\:\mathrm{4}{x}}−\frac{\mathrm{8}}{\:\sqrt{\mathrm{5}}}\int\frac{{dx}}{\mathrm{3}+\frac{\mathrm{4}}{\:\sqrt{\mathrm{5}}}+\mathrm{cos}\:\mathrm{4}{x}}= \\ $$$$\left[\mathrm{using}\:{t}=\mathrm{tan}\:\mathrm{2}{x}\right] \\ $$$$=\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{5}}\right)\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}+\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{5}}\right)\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}= \\ $$$$=−\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\frac{\left(\mathrm{1}−\sqrt{\mathrm{5}}\right){t}}{\mathrm{4}}\:+\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right){t}}{\mathrm{4}}\right)= \\ $$$$=−\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\frac{\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)\mathrm{tan}\:\mathrm{2}{x}}{\mathrm{4}}\:+\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{tan}\:\mathrm{2}{x}}{\mathrm{4}}\right)+{C} \\ $$