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Question Number 139285 by mohammad17 last updated on 25/Apr/21

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{6}} \left({x}\right){dx} \\$$

Answered by qaz last updated on 25/Apr/21

$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{6}} {xdx}=\frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{7}}{\mathrm{2}}\right) \\$$ $$=\frac{\frac{\mathrm{1}×\mathrm{5}×\mathrm{3}×\mathrm{1}}{\mathrm{2}×\mathrm{2}×\mathrm{2}×\mathrm{2}}\pi}{\mathrm{2}×\mathrm{4}×\mathrm{3}×\mathrm{2}}=\frac{\mathrm{5}\pi}{\mathrm{256}} \\$$

Commented bymohammad17 last updated on 25/Apr/21

$${thank}\:{you}\:{sir}\:{can}\:{you}\:{give}\:{me}\:{steb}\:{by}\:{steb} \\$$ $${please}? \\$$

Commented byqaz last updated on 25/Apr/21

$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}\:^{{p}} {x}\mathrm{cos}\:^{{q}} {xdx}=\frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{{p}+\mathrm{1}}{\mathrm{2}},\frac{{q}+\mathrm{1}}{\mathrm{2}}\right) \\$$ $$=\frac{\Gamma\left(\frac{{p}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{{q}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{{p}+{q}+\mathrm{2}}{\mathrm{2}}\right)} \\$$ $$\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right) \\$$ $$\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi} \\$$

Answered by MJS_new last updated on 25/Apr/21

$$\int\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{6}} \:{x}\:{dx}= \\$$ $$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}\right] \\$$ $$=\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{5}} }{dt}= \\$$ $$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\$$ $$=\frac{{t}\left(\mathrm{15}{t}^{\mathrm{6}} +\mathrm{55}{t}^{\mathrm{4}} +\mathrm{73}{t}^{\mathrm{2}} −\mathrm{15}\right)}{\mathrm{384}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }+\frac{\mathrm{5}}{\mathrm{128}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}= \\$$ $$=\frac{{t}\left(\mathrm{15}{t}^{\mathrm{6}} +\mathrm{55}{t}^{\mathrm{4}} +\mathrm{73}{t}^{\mathrm{2}} −\mathrm{15}\right)}{\mathrm{384}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }+\frac{\mathrm{5}}{\mathrm{128}}\mathrm{arctan}\:{t} \\$$ $$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{5}\pi}{\mathrm{256}} \\$$