Question Number 55560 by Tawa1 last updated on 26/Feb/19
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Feb/19
$${x}={tana}\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sec}^{\mathrm{6}} {a}×{sec}^{\mathrm{2}} {ada}}{\left({tan}^{\mathrm{4}} {a}+{sec}^{\mathrm{2}} {a}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} } \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{da}}{{cos}^{\mathrm{8}} {a}\left\{\frac{{sin}^{\mathrm{4}} {a}}{{cos}^{\mathrm{4}} {a}}+\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {a}}\right\}^{\frac{\mathrm{5}}{\mathrm{2}}} } \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{da}}{{cos}^{\mathrm{8}} {a}\left\{\frac{{sin}^{\mathrm{4}} {a}+{cos}^{\mathrm{2}} {a}}{{cos}^{\mathrm{4}} {a}}\right\}^{\frac{\mathrm{5}}{\mathrm{2}}} } \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{10}} {a}}{{cos}^{\mathrm{8}} {a}\left({sin}^{\mathrm{4}} {a}+{cos}^{\mathrm{2}} {a}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{da} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{2}} {ada}}{\left({sin}^{\mathrm{4}} {a}+{cos}^{\mathrm{2}} {a}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{da} \\ $$$${wait}… \\ $$