Question Number 52164 by pooja24 last updated on 04/Jan/19
$${find}: \\ $$$$\underset{\mathrm{0}} {\overset{\Pi} {\int}}\left({cos}^{\mathrm{6}} \theta\:−\mathrm{cos}\:^{\mathrm{4}} \theta\right)\:{d}\theta \\ $$$${plase}\:{help}\:{me}\:{in}\:{cinding}\:{this}\:{And}\:{also} \\ $$$${explain}\:{if}\:{possible} \\ $$
Commented by prakash jain last updated on 04/Jan/19
$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}^{{m}} {x}\mathrm{cos}^{{n}} {xdx} \\ $$$$=\frac{\left.\left[\left.\left({n}−\mathrm{1}\right)\left({n}−\mathrm{3}\right)..\mathrm{1}\:\mathrm{or}\:\mathrm{2}\right]\right]\left({m}−\mathrm{1}\right)\left({m}−\mathrm{3}\right)..\mathrm{1}{or}\mathrm{2}\right]}{\left({m}+{n}\right)\left({m}+{n}−\mathrm{2}\right)…\:\mathrm{1}\:\mathrm{or}\:\mathrm{2}}\mathrm{K} \\ $$$$\mathrm{where}{K}=\frac{\pi}{\mathrm{2}}\:\mathrm{if}\:\mathrm{both}\:{m},{n}\:\mathrm{are}\:\mathrm{even} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{K}=\mathrm{1}\:\mathrm{otherwise} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{6}} {xdx}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{4}} {xdx} \\ $$$$+\int_{\pi/\mathrm{2}} ^{\pi} \mathrm{cos}^{\mathrm{6}} {xdx}−\int_{\pi/\mathrm{2}} ^{\pi} \mathrm{cos}^{\mathrm{4}} {xdx} \\ $$$$\int_{\pi/\mathrm{2}} ^{\pi} \mathrm{cos}^{\mathrm{6}} {xdx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}^{\mathrm{6}} {xdx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{6}} {xdx} \\ $$$$\int_{\pi/\mathrm{2}} ^{\pi} \mathrm{cos}^{\mathrm{4}} {xdx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}^{\mathrm{4}} {xdx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{4}} {xdx} \\ $$$$\mathrm{I}=\mathrm{2}\left[\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{6}} {xdx}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{4}} {xdx}\right] \\ $$$$=\mathrm{2}\left[\frac{\mathrm{5}\centerdot\mathrm{3}\centerdot\mathrm{1}}{\mathrm{6}\centerdot\mathrm{4}\centerdot\mathrm{2}}−\frac{\mathrm{3}\centerdot\mathrm{1}}{\mathrm{4}\centerdot\mathrm{2}}\right]\frac{\pi}{\mathrm{2}}\:\:\because{m}=\mathrm{6}\:\mathrm{even} \\ $$$$=\mathrm{2}\left[\frac{\mathrm{5}}{\mathrm{16}}−\frac{\mathrm{3}}{\mathrm{8}}\right]\frac{\pi}{\mathrm{2}} \\ $$$$=−\frac{\pi}{\mathrm{16}} \\ $$
Commented by maxmathsup by imad last updated on 04/Jan/19
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\pi} \left({cos}^{\mathrm{6}} {x}−{cos}^{\mathrm{4}} {x}\right){dx}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\pi} \left({cos}^{\mathrm{4}} {x}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{4}} {x}\right){dx}\right. \\ $$$$=−\int_{\mathrm{0}} ^{\pi} \:{sin}^{\mathrm{2}} {x}\:{cos}^{\mathrm{4}} {x}\:{dx}\:=−\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right)^{\mathrm{2}} {dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}−{cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right)\left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}−\frac{\mathrm{1}+{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}\right)\left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\pi} \left(\frac{\mathrm{1}−{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}\right)\left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{16}}\:\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)−{cos}\left(\mathrm{4}{x}\right)−{cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{4}{x}\right)\right){dx} \\ $$$$=−\frac{\pi}{\mathrm{16}}\:−\frac{\mathrm{1}}{\mathrm{16}}\:\int_{\mathrm{0}} ^{\pi} {cos}\left(\mathrm{2}{x}\right){dx}+\frac{\mathrm{1}}{\mathrm{16}}\:\int_{\mathrm{0}} ^{\pi} {cos}\left(\mathrm{4}{x}\right){dx}\:+\frac{\mathrm{1}}{\mathrm{16}}\:\int_{\mathrm{0}} ^{\pi} {cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{4}{x}\right){dx} \\ $$$$=−\frac{\pi}{\mathrm{16}}\:−\frac{\mathrm{1}}{\mathrm{32}}\left[{sin}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{\pi} \:+\frac{\mathrm{1}}{\mathrm{64}}\left[{sin}\left(\mathrm{4}{x}\right)\right]_{\mathrm{0}} ^{\pi} \:\:\:+\frac{\mathrm{1}}{\mathrm{32}}\:\int_{\mathrm{0}} ^{\pi} \left({cos}\left(\mathrm{6}{x}\right)+{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=−\frac{\pi}{\mathrm{16}}\:+\mathrm{0}\:+\mathrm{0}\:+\mathrm{0}\:=−\frac{\pi}{\mathrm{16}}\:. \\ $$
Commented by maxmathsup by imad last updated on 04/Jan/19
$${error}\:{of}\:{typo}\:{at}\:{first}\:{line}\:{I}\:=\int_{\mathrm{0}} ^{\pi} \left\{{cos}^{\mathrm{4}} {x}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)−{cos}^{\mathrm{4}} {x}\right\}{dx}… \\ $$
Commented by pooja24 last updated on 05/Jan/19
$${thank}\:{you} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jan/19
$$−{cos}^{\mathrm{4}} \theta\left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right) \\ $$$$=−{cos}^{\mathrm{4}} \theta{sin}^{\mathrm{2}} \theta \\ $$$$=−\left(\frac{\mathrm{1}+{cos}\mathrm{2}\theta}{\mathrm{2}}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}−{cos}\mathrm{2}\theta}{\mathrm{2}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+{cos}\mathrm{2}\theta\right)\left(\mathrm{1}−{cos}^{\mathrm{2}} \mathrm{2}\theta\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+{cos}\mathrm{2}\theta\right)\left(\mathrm{1}−\frac{\mathrm{1}+{cos}\mathrm{4}\theta}{\mathrm{2}}\right) \\ $$$$=\frac{−\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}+{cos}\mathrm{2}\theta\right)\left(\mathrm{2}−\mathrm{1}−{cos}\mathrm{4}\theta\right) \\ $$$$=\frac{−\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}+{cos}\mathrm{2}\theta\right)\left(\mathrm{1}−{cos}\mathrm{4}\theta\right) \\ $$$$=\frac{−\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}−{cos}\mathrm{4}\theta+{cos}\mathrm{2}\theta−{cos}\mathrm{2}\theta{cos}\mathrm{4}\theta\right) \\ $$$$=\frac{−\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}−{cos}\mathrm{4}\theta+{cos}\mathrm{2}\theta−\frac{{cos}\mathrm{6}\theta+{cos}\mathrm{2}\theta}{\mathrm{2}}\right) \\ $$$$=\frac{−\mathrm{1}}{\mathrm{32}}\left(\mathrm{2}−\mathrm{2}{cos}\mathrm{4}\theta+\mathrm{2}{cos}\mathrm{2}\theta−{cos}\mathrm{6}\theta−{cos}\mathrm{2}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\left({cos}\mathrm{6}\theta+\mathrm{2}{cos}\mathrm{4}\theta−{cos}\mathrm{2}\theta−\mathrm{2}\right) \\ $$$${now}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{32}}\int_{\mathrm{0}} ^{\pi} {cos}\mathrm{6}\theta+\mathrm{2}{cos}\mathrm{4}\theta−{cos}\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\pi} {d}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{32}}\mid\frac{{sin}\mathrm{6}\theta}{\mathrm{6}}+\frac{\mathrm{2}{sin}\mathrm{4}\theta}{\mathrm{4}}−\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\mid_{\mathrm{0}} ^{\pi} −\frac{\mathrm{1}}{\mathrm{16}}×\pi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{} {\:}\:\underset{{intregal}\:{value}\:{zer}\underset{} {{o}}} {\Downarrow}\:\:\:\:−\frac{\mathrm{1}}{\mathrm{16}}×\pi \\ $$$$ \\ $$$${answer}\:{is}\:=\frac{−\pi}{\mathrm{16}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{sinn}\pi=\mathrm{0}\right] \\ $$$${pls}\:{chek}… \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jan/19