Question Number 48498 by maxmathsup by imad last updated on 24/Nov/18
$${find}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{{n}} {xdx}\:\:{and}\:{B}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{sin}^{{n}} {xdx} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{\mathrm{6}} {xdx}\:{and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{sin}^{\mathrm{6}} {xdx}\:. \\ $$
Commented by Abdulhafeez Abu qatada last updated on 24/Nov/18
$$ \\ $$$$ \\ $$$$ \\ $$$${i}\:{made}\:{an}\:{error},\:{but}\:{its}\:{corrected} \\ $$
Commented by maxmathsup by imad last updated on 25/Nov/18
$$\left.\mathrm{1}\right)\:{W}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left(\frac{{e}^{{ix}} \:+{e}^{−{ix}} }{\mathrm{2}}\right)^{{n}} {dx}\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left({e}^{{ix}} \:+{e}^{−{ix}} \right)^{{n}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left({e}^{{ix}} \right)^{{k}} \:\left({e}^{−{ix}} \right)^{{n}−{k}} \right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:{e}^{{ikx}} \:{e}^{−{i}\left({n}−{k}\right){x}} {dx}\:\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{A}_{{k}} \\ $$$${A}_{{k}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{\left({ik}−{in}\:+{ik}\right){x}} {dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:{e}^{\left(\mathrm{2}{ik}−{in}\right){x}} {dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:{e}^{{i}\left(\mathrm{2}{k}−{n}\right){x}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{k}−{n}\right){x}\:{dx}\:+{i}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{sin}\left(\mathrm{2}{k}−{n}\right){x}\:{dx} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}{k}−{n}}{sin}\left(\mathrm{2}{k}−{n}\right){x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:−{i}\:\left[\frac{\mathrm{1}}{\mathrm{2}{k}−{n}}\:{cos}\left(\mathrm{2}{k}−{n}\right){x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{k}−{n}}{sin}\left(\mathrm{2}{k}−{n}\right)\frac{\pi}{\mathrm{4}}\:\:+{i}\:\left(\frac{\mathrm{1}}{\mathrm{2}{k}−{n}}\:−\frac{\mathrm{1}}{\mathrm{2}{k}−{n}}\:{cos}\left(\mathrm{2}{k}−{n}\right)\frac{\pi}{\mathrm{4}}\right)\:\Rightarrow \\ $$$${W}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \frac{\mathrm{1}}{\mathrm{2}{k}−{n}}\left\{{sin}\left(\mathrm{2}{k}−{n}\right)\frac{\pi}{\mathrm{4}}\:+{i}\left(\mathrm{1}−{cos}\left(\mathrm{2}{k}−{n}\right)\frac{\pi}{\mathrm{4}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:{C}_{{n}} ^{{k}} \:\:\frac{{sin}\left(\mathrm{2}{k}−{n}\right)\frac{\pi}{\mathrm{4}}}{\mathrm{2}{k}−{n}}\:+{i}\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{k}−{n}\right)\frac{\pi}{\mathrm{4}}}{\mathrm{2}{k}−{n}}\:{but}\:{W}_{{n}} {is}\:{real}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{k}−{n}\right)\frac{\pi}{\mathrm{4}}}{\mathrm{2}{k}−{n}}\:\:=\mathrm{0}\:{and}\:{W}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\frac{{sin}\left(\mathrm{2}{k}−{n}\right)\frac{\pi}{\mathrm{4}}}{\mathrm{2}{k}−{n}}\:. \\ $$
Commented by maxmathsup by imad last updated on 25/Nov/18
$${we}\:{have}\:{B}_{{n}\:} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left(\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}\right)^{{n}} {dx}\:=\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{{n}} }\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({e}^{{ix}} \right)^{{k}} \left(−\mathrm{1}\right)^{{n}−{k}} \:\left({e}^{−{ix}} \right)^{{n}−{k}} {dx}\right. \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{n}−{k}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{{ikx}} \:{e}^{−{i}\left({n}−{k}\right){x}} {dx} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\left(−\mathrm{1}\right)^{{n}−{k}} \:{C}_{{n}} ^{{k}} \:\:.{A}_{{k}} \:\:\:{with}\:{A}_{{k}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:{e}^{\left({ik}−{in}+{ik}\right){x}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{{i}\left(\mathrm{2}{k}−{n}\right){x}} {dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{k}−{n}\right){x}\:{dx}\:+{i}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{sin}\left(\mathrm{2}{k}−{n}\right){x}\:{dx} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}{k}−{n}}{sin}\left(\mathrm{2}{k}−{n}\right){x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:+{i}\left[−\frac{\mathrm{1}}{\mathrm{2}{k}−{n}}{cos}\left(\mathrm{2}{k}−{n}\right){x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{k}−{n}}{sin}\left(\mathrm{2}{k}−{n}\right)\frac{\pi}{\mathrm{4}}\:+\frac{{i}}{\mathrm{2}{k}−{n}}\left(\:\mathrm{1}−{cos}\left(\mathrm{2}{k}−{n}\right)\frac{\pi}{\mathrm{4}}\right)\:\Rightarrow \\ $$$${B}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{n}−{k}} \:{C}_{{n}} ^{{k}} \:\frac{\mathrm{1}}{\mathrm{2}{k}−{n}}\left\{\:{sin}\left(\mathrm{2}{k}−{n}\right)\frac{\pi}{\mathrm{4}}\:+{i}\left(\mathrm{1}−{cos}\left(\mathrm{2}{k}−{n}\right)\frac{\pi}{\mathrm{4}}\right)\right\} \\ $$$$\:\Rightarrow{B}_{\mathrm{2}{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} \left(−\mathrm{1}\right)^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:\left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{2}{n}} ^{{k}} \:\:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{2}{n}}\left\{{sin}\left({k}−{n}\right)\frac{\pi}{\mathrm{2}}\:+{i}\left(\mathrm{1}−{cos}\left({k}−{n}\right)\frac{\pi}{\mathrm{2}}\right)\right\} \\ $$$${but}\:{B}_{\mathrm{2}{n}} \:\:{is}\:{real}\:\Rightarrow\:{B}_{\mathrm{2}{n}} \:=\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:\left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{2}{n}} ^{{k}} \:\:\:\frac{{sin}\left({k}−{n}\right)\frac{\pi}{\mathrm{2}}}{{k}−{n}}\: \\ $$$${and}\:{we}\:{follow}\:{the}\:{same}\:{method}\:{to}\:{find}\:{B}_{\mathrm{2}{n}+\mathrm{1}} \\ $$
Commented by maxmathsup by imad last updated on 25/Nov/18
$$\left.\mathrm{2}\right)\:{let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{\mathrm{6}} {x}\:{dx}\:{and}\:{J}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{sin}^{\mathrm{6}} {x}\:{dx}\:\Rightarrow \\ $$$${I}\:+{J}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left\{\:\left({cos}^{\mathrm{2}} {x}\right)^{\mathrm{3}\:} \:+\left({sin}^{\mathrm{2}} {x}\right)^{\mathrm{3}} \right\}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left\{\:\left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{3}} −\mathrm{3}{cos}^{\mathrm{2}} {x}\:{sin}^{\mathrm{2}} {x}\left({cos}^{\mathrm{2}} {x}\:+{sin}^{\mathrm{2}} {x}\right)\right\}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left\{\mathrm{1}−\mathrm{3}\:{cos}^{\mathrm{2}} {x}\:{sin}^{\mathrm{2}} {x}\right\}{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left\{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)\right)\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right){dx}\right. \\ $$$$=\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{3}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left(\mathrm{1}−{cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{3}}{\mathrm{4}}.\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{3}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{1}+{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}{dx} \\ $$$$=\frac{\pi}{\mathrm{16}}\:\:+\frac{\mathrm{3}}{\mathrm{8}}.\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{3}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{4}{x}\right){dx} \\ $$$$=\frac{\pi}{\mathrm{16}}\:+\frac{\mathrm{3}\pi}{\mathrm{32}}\:+\mathrm{0}\:=\frac{\mathrm{5}\pi}{\mathrm{32}}\:\:{also}\:{we}\:{have}\:{I}−{J}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left(\:\left({cos}^{\mathrm{2}} {x}\right)^{\mathrm{3}} −\left({sin}^{\mathrm{2}} {x}\right)^{\mathrm{3}} {dx}\right. \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left({cos}^{\mathrm{2}} {x}\:−{sin}^{\mathrm{2}} {x}\right)\left({cos}^{\mathrm{4}} {x}\:+{cos}^{\mathrm{2}} {x}\:{sin}^{\mathrm{2}} {x}\:+{sin}^{\mathrm{4}} {x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {cos}\left(\mathrm{2}{x}\right)\left(\:\left({cos}^{\mathrm{2}} {x}\:+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −{cos}^{\mathrm{2}} {x}\:{sin}^{\mathrm{2}} {x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{x}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right){dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{x}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}−{cos}\left(\mathrm{4}{x}\right)\right){dx}\right. \\ $$$$=\frac{\mathrm{7}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{x}\right){dx}\:+\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{4}{x}\right){dx}\:{but}\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{x}\right){dx}\:=\left[\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{4}{x}\right)\:{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left({cos}\left(\mathrm{6}{x}\right)+{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left[{sin}\left(\mathrm{6}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:=\frac{\mathrm{1}}{\mathrm{12}}{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\:=−\frac{\mathrm{1}}{\mathrm{12}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:=\frac{−\mathrm{1}+\mathrm{3}}{\mathrm{12}}\:=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow\:{I}\:−{J}\:=\frac{\mathrm{7}}{\mathrm{16}}\:+\frac{\mathrm{1}}{\mathrm{48}}\:=\frac{\mathrm{22}}{\mathrm{48}}\:=\frac{\mathrm{11}}{\mathrm{24}}\:\Rightarrow\:{I}\:+{J}\:=\frac{\mathrm{5}\pi}{\mathrm{32}}\:{and}\:{I}\:−{J}\:=\frac{\mathrm{11}}{\mathrm{24}}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\frac{\mathrm{5}\pi}{\mathrm{32}}\:+\frac{\mathrm{11}}{\mathrm{24}}\:\Rightarrow\:{I}\:=\frac{\mathrm{5}\pi}{\mathrm{64}}\:+\frac{\mathrm{11}}{\mathrm{48}}\:{also}\:\:\mathrm{2}{J}\:=\frac{\mathrm{5}\pi}{\mathrm{32}}\:−\frac{\mathrm{11}}{\mathrm{24}}\:\Rightarrow\:{J}\:=\frac{\mathrm{5}\pi}{\mathrm{64}}\:−\frac{\mathrm{11}}{\mathrm{48}}\:. \\ $$$${J} \\ $$$$ \\ $$
Answered by Abdulhafeez Abu qatada last updated on 24/Nov/18
$$ \\ $$$$ \\ $$$${If}\:{I}_{{n}} \:=\:\int{sin}^{{n}} {x}\:{dx},\:{I}_{{n}} \:=\:\frac{−\mathrm{cos}{x}.\mathrm{sin}^{{n}−\mathrm{1}} {x}}{{n}}\:+\:\frac{{n}−\mathrm{1}}{{n}}{I}_{{n}−\mathrm{2}} \\ $$$$\therefore\:{B}_{{n}} \:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}{sin}^{{n}} {x}\:{dx},\:{B}_{{n}} \:=\:\left[\frac{−\mathrm{cos}{x}.\mathrm{sin}^{{n}−\mathrm{1}} {x}}{{n}}\:+\:\frac{{n}−\mathrm{1}}{{n}}{A}_{{n}−\mathrm{2}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$${B}_{{n}} \:=\:\frac{−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{{n}} }{{n}}\:+\:\frac{{n}−\mathrm{1}}{{n}}{B}_{{n}−\mathrm{2}} \: \\ $$$$ \\ $$$${A}_{{n}} \:=\:\left[\frac{\mathrm{sin}{x}.\mathrm{cos}^{{n}−\mathrm{1}} {x}}{{n}}\:+\:\frac{{n}−\mathrm{1}}{{n}}{A}_{{n}−\mathrm{2}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$${A}_{{n}} \:=\:\frac{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{{n}} }{{n}}\:+\:\frac{{n}−\mathrm{1}}{{n}}{A}_{{n}−\mathrm{2}} \: \\ $$$${A}_{\mathrm{6}} \:=\:\frac{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{6}} }{\mathrm{6}}\:+\:\frac{\mathrm{5}}{\mathrm{6}}{A}_{\mathrm{4}} \: \\ $$$${A}_{\mathrm{6}} \:=\:\frac{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{6}} }{\mathrm{6}}\:+\:\frac{\mathrm{5}}{\mathrm{6}}\left(\frac{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{4}} }{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}{A}_{\mathrm{2}} \right) \\ $$$${A}_{\mathrm{6}} \:=\:\frac{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{6}} }{\mathrm{6}}\:+\:\frac{\mathrm{5}}{\mathrm{6}}\left(\frac{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{4}} }{\mathrm{4}}\:\right)\:+\:\frac{\mathrm{5}}{\mathrm{6}}.\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{A}_{\mathrm{0}} \right) \\ $$$${A}_{\mathrm{0}} \:=\:\underset{\mathrm{0}} {\int}^{\frac{\pi}{\mathrm{4}}} \:{dx}\:=\:\frac{\pi}{\mathrm{4}} \\ $$$${A}_{\mathrm{6}} \:=\:\frac{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{6}} }{\mathrm{6}}\:+\:\frac{\mathrm{5}}{\mathrm{6}}\left(\frac{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{4}} }{\mathrm{4}}\:\right)\:+\:\frac{\mathrm{5}}{\mathrm{6}}.\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}\:+\:\frac{\pi}{\mathrm{8}}\right) \\ $$$$ \\ $$$${A}_{\mathrm{6}} \:=\:\:\frac{\mathrm{5}\pi}{\mathrm{64}}\:+\:\frac{\mathrm{11}}{\mathrm{48}}…..{After}\:{simplifying} \\ $$$$ \\ $$$${B}_{\mathrm{6}} \:=\:\frac{−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{6}} }{\mathrm{6}}\:+\:\frac{\mathrm{5}}{\mathrm{6}}{B}_{\mathrm{4}} \: \\ $$$${B}_{\mathrm{6}} \:=\:\frac{−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{6}} }{\mathrm{6}}\:+\:\frac{\mathrm{5}}{\mathrm{6}}\left(\frac{−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{4}} }{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}{B}_{\mathrm{2}} \right) \\ $$$${B}_{\mathrm{6}} \:=\:\frac{−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{6}} }{\mathrm{6}}\:+\:\frac{\mathrm{5}}{\mathrm{6}}\left(\frac{−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{4}} }{\mathrm{4}}\:\right)\:+\:\frac{\mathrm{5}}{\mathrm{6}}.\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{B}_{\mathrm{0}} \right) \\ $$$$ \\ $$$${B}_{\mathrm{0}} \:=\:\underset{\mathrm{0}} {\int}^{\frac{\pi}{\mathrm{4}}} \:{dx}\:=\:\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$$${B}_{\mathrm{6}} \:=\:\frac{−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{6}} }{\mathrm{6}}\:+\:\frac{\mathrm{5}}{\mathrm{6}}\left(\frac{−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{4}} }{\mathrm{4}}\:\right)\:+\:\frac{\mathrm{5}}{\mathrm{6}}.\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}\:+\:\frac{\pi}{\mathrm{8}}\right) \\ $$$$ \\ $$$${B}_{\mathrm{6}} \:=\:\frac{\mathrm{5}\pi}{\mathrm{64}}\:−\:\frac{\mathrm{11}}{\mathrm{48}}……{After}\:{simplifying} \\ $$$$ \\ $$$${Abu}\:{qatada} \\ $$
Answered by Abdulhafeez Abu qatada last updated on 24/Nov/18