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dx-sin-5-x-cos-7-x-




Question Number 126502 by benjo_mathlover last updated on 21/Dec/20
  ∫ (dx/( (√(sin^5 x cos^7 x)))) ?
$$\:\:\int\:\frac{{dx}}{\:\sqrt{\mathrm{sin}\:^{\mathrm{5}} {x}\:\mathrm{cos}\:^{\mathrm{7}} {x}}}\:? \\ $$
Answered by MJS_new last updated on 21/Dec/20
∫(dx/( (√(sin^5  x cos^7  x))))=       [t=(√(tan x)) → dx=2cos^2  x (√(tan x)) dt]  =2∫(((t^4 +1)^2 )/t^4 )dt=∫(2t^4 +4+(2/t^4 ))dt=  =(2/5)t^5 +4t−(2/(3t^3 ))=...
$$\int\frac{{dx}}{\:\sqrt{\mathrm{sin}^{\mathrm{5}} \:{x}\:\mathrm{cos}^{\mathrm{7}} \:{x}}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\mathrm{2cos}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}}\:{dt}\right] \\ $$$$=\mathrm{2}\int\frac{\left({t}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{4}} }{dt}=\int\left(\mathrm{2}{t}^{\mathrm{4}} +\mathrm{4}+\frac{\mathrm{2}}{{t}^{\mathrm{4}} }\right){dt}= \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}{t}^{\mathrm{5}} +\mathrm{4}{t}−\frac{\mathrm{2}}{\mathrm{3}{t}^{\mathrm{3}} }=… \\ $$

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