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
![∫_0 ^(π/2) sin^m xcos^n xdx =(([(n−1)(n−3)..1 or 2]](m−1)(m−3)..1or2])/((m+n)(m+n−2)... 1 or 2))K whereK=(π/2) if both m,n are even K=1 otherwise I=∫_0 ^(π/2) cos^6 xdx−∫_0 ^(π/2) cos^4 xdx +∫_(π/2) ^π cos^6 xdx−∫_(π/2) ^π cos^4 xdx ∫_(π/2) ^π cos^6 xdx=∫_0 ^(π/2) sin^6 xdx=∫_0 ^(π/2) cos^6 xdx ∫_(π/2) ^π cos^4 xdx=∫_0 ^(π/2) sin^4 xdx=∫_0 ^(π/2) cos^4 xdx I=2[∫_0 ^(π/2) cos^6 xdx−∫_0 ^(π/2) cos^4 xdx] =2[((5∙3∙1)/(6∙4∙2))−((3∙1)/(4∙2))](π/2) ∵m=6 even =2[(5/(16))−(3/8)](π/2) =−(π/(16))](https://www.tinkutara.com/question/Q52179.png)
$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}^{{m}} {x}\mathrm{cos}^{{n}} {xdx} \\ $$$$=\frac{\left.\left[\left.\left({n}−\mathrm{1}\right)\left({n}−\mathrm{3}\right)..\mathrm{1}\:\mathrm{or}\:\mathrm{2}\right]\right]\left({m}−\mathrm{1}\right)\left({m}−\mathrm{3}\right)..\mathrm{1}{or}\mathrm{2}\right]}{\left({m}+{n}\right)\left({m}+{n}−\mathrm{2}\right)…\:\mathrm{1}\:\mathrm{or}\:\mathrm{2}}\mathrm{K} \\ $$$$\mathrm{where}{K}=\frac{\pi}{\mathrm{2}}\:\mathrm{if}\:\mathrm{both}\:{m},{n}\:\mathrm{are}\:\mathrm{even} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{K}=\mathrm{1}\:\mathrm{otherwise} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{6}} {xdx}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{4}} {xdx} \\ $$$$+\int_{\pi/\mathrm{2}} ^{\pi} \mathrm{cos}^{\mathrm{6}} {xdx}−\int_{\pi/\mathrm{2}} ^{\pi} \mathrm{cos}^{\mathrm{4}} {xdx} \\ $$$$\int_{\pi/\mathrm{2}} ^{\pi} \mathrm{cos}^{\mathrm{6}} {xdx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}^{\mathrm{6}} {xdx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{6}} {xdx} \\ $$$$\int_{\pi/\mathrm{2}} ^{\pi} \mathrm{cos}^{\mathrm{4}} {xdx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}^{\mathrm{4}} {xdx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{4}} {xdx} \\ $$$$\mathrm{I}=\mathrm{2}\left[\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{6}} {xdx}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{4}} {xdx}\right] \\ $$$$=\mathrm{2}\left[\frac{\mathrm{5}\centerdot\mathrm{3}\centerdot\mathrm{1}}{\mathrm{6}\centerdot\mathrm{4}\centerdot\mathrm{2}}−\frac{\mathrm{3}\centerdot\mathrm{1}}{\mathrm{4}\centerdot\mathrm{2}}\right]\frac{\pi}{\mathrm{2}}\:\:\because{m}=\mathrm{6}\:\mathrm{even} \\ $$$$=\mathrm{2}\left[\frac{\mathrm{5}}{\mathrm{16}}−\frac{\mathrm{3}}{\mathrm{8}}\right]\frac{\pi}{\mathrm{2}} \\ $$$$=−\frac{\pi}{\mathrm{16}} \\ $$
Commented by maxmathsup by imad last updated on 04/Jan/19
![let I =∫_0 ^π (cos^6 x−cos^4 x)dx ⇒I =∫_0 ^π (cos^4 x(1−sin^2 x−cos^4 x)dx =−∫_0 ^π sin^2 x cos^4 x dx =−∫_0 ^π ((1−cos(2x))/2)(((1+cos(2x))/2))^2 dx =−(1/8) ∫_0 ^π (1−cos^2 (2x))(1+cos(2x))dx =−(1/8) ∫_0 ^π (1−((1+cos(4x))/2))(1+cos(2x))dx =−(1/8) ∫_0 ^π (((1−cos(4x))/2))(1+cos(2x))dx =−(1/(16)) ∫_0 ^π (1+cos(2x)−cos(4x)−cos(2x)cos(4x))dx =−(π/(16)) −(1/(16)) ∫_0 ^π cos(2x)dx+(1/(16)) ∫_0 ^π cos(4x)dx +(1/(16)) ∫_0 ^π cos(2x)cos(4x)dx =−(π/(16)) −(1/(32))[sin(2x)]_0 ^π +(1/(64))[sin(4x)]_0 ^π +(1/(32)) ∫_0 ^π (cos(6x)+cos(2x))dx =−(π/(16)) +0 +0 +0 =−(π/(16)) .](https://www.tinkutara.com/question/Q52195.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\pi} \left({cos}^{\mathrm{6}} {x}−{cos}^{\mathrm{4}} {x}\right){dx}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\pi} \left({cos}^{\mathrm{4}} {x}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{4}} {x}\right){dx}\right. \\ $$$$=−\int_{\mathrm{0}} ^{\pi} \:{sin}^{\mathrm{2}} {x}\:{cos}^{\mathrm{4}} {x}\:{dx}\:=−\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right)^{\mathrm{2}} {dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}−{cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right)\left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}−\frac{\mathrm{1}+{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}\right)\left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\pi} \left(\frac{\mathrm{1}−{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}\right)\left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{16}}\:\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)−{cos}\left(\mathrm{4}{x}\right)−{cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{4}{x}\right)\right){dx} \\ $$$$=−\frac{\pi}{\mathrm{16}}\:−\frac{\mathrm{1}}{\mathrm{16}}\:\int_{\mathrm{0}} ^{\pi} {cos}\left(\mathrm{2}{x}\right){dx}+\frac{\mathrm{1}}{\mathrm{16}}\:\int_{\mathrm{0}} ^{\pi} {cos}\left(\mathrm{4}{x}\right){dx}\:+\frac{\mathrm{1}}{\mathrm{16}}\:\int_{\mathrm{0}} ^{\pi} {cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{4}{x}\right){dx} \\ $$$$=−\frac{\pi}{\mathrm{16}}\:−\frac{\mathrm{1}}{\mathrm{32}}\left[{sin}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{\pi} \:+\frac{\mathrm{1}}{\mathrm{64}}\left[{sin}\left(\mathrm{4}{x}\right)\right]_{\mathrm{0}} ^{\pi} \:\:\:+\frac{\mathrm{1}}{\mathrm{32}}\:\int_{\mathrm{0}} ^{\pi} \left({cos}\left(\mathrm{6}{x}\right)+{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=−\frac{\pi}{\mathrm{16}}\:+\mathrm{0}\:+\mathrm{0}\:+\mathrm{0}\:=−\frac{\pi}{\mathrm{16}}\:. \\ $$
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^4 θ(1−cos^2 θ) =−cos^4 θsin^2 θ =−(((1+cos2θ)/2))^2 (((1−cos2θ)/2)) =−(1/8)(1+cos2θ)(1−cos^2 2θ) =−(1/8)(1+cos2θ)(1−((1+cos4θ)/2)) =((−1)/(16))(1+cos2θ)(2−1−cos4θ) =((−1)/(16))(1+cos2θ)(1−cos4θ) =((−1)/(16))(1−cos4θ+cos2θ−cos2θcos4θ) =((−1)/(16))(1−cos4θ+cos2θ−((cos6θ+cos2θ)/2)) =((−1)/(32))(2−2cos4θ+2cos2θ−cos6θ−cos2θ) =(1/(32))(cos6θ+2cos4θ−cos2θ−2) now (1/(32))∫_0 ^π cos6θ+2cos4θ−cos2θ−(1/(16))∫_0 ^π dθ (1/(32))∣((sin6θ)/6)+((2sin4θ)/4)−((sin2θ)/2)∣_0 ^π −(1/(16))×π _ ⇓_(intregal value zero_ ) −(1/(16))×π answer is =((−π)/(16)) [sinnπ=0] pls chek...](https://www.tinkutara.com/question/Q52170.png)
$$−{cos}^{\mathrm{4}} \theta\left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right) \\ $$$$=−{cos}^{\mathrm{4}} \theta{sin}^{\mathrm{2}} \theta \\ $$$$=−\left(\frac{\mathrm{1}+{cos}\mathrm{2}\theta}{\mathrm{2}}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}−{cos}\mathrm{2}\theta}{\mathrm{2}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+{cos}\mathrm{2}\theta\right)\left(\mathrm{1}−{cos}^{\mathrm{2}} \mathrm{2}\theta\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+{cos}\mathrm{2}\theta\right)\left(\mathrm{1}−\frac{\mathrm{1}+{cos}\mathrm{4}\theta}{\mathrm{2}}\right) \\ $$$$=\frac{−\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}+{cos}\mathrm{2}\theta\right)\left(\mathrm{2}−\mathrm{1}−{cos}\mathrm{4}\theta\right) \\ $$$$=\frac{−\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}+{cos}\mathrm{2}\theta\right)\left(\mathrm{1}−{cos}\mathrm{4}\theta\right) \\ $$$$=\frac{−\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}−{cos}\mathrm{4}\theta+{cos}\mathrm{2}\theta−{cos}\mathrm{2}\theta{cos}\mathrm{4}\theta\right) \\ $$$$=\frac{−\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}−{cos}\mathrm{4}\theta+{cos}\mathrm{2}\theta−\frac{{cos}\mathrm{6}\theta+{cos}\mathrm{2}\theta}{\mathrm{2}}\right) \\ $$$$=\frac{−\mathrm{1}}{\mathrm{32}}\left(\mathrm{2}−\mathrm{2}{cos}\mathrm{4}\theta+\mathrm{2}{cos}\mathrm{2}\theta−{cos}\mathrm{6}\theta−{cos}\mathrm{2}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\left({cos}\mathrm{6}\theta+\mathrm{2}{cos}\mathrm{4}\theta−{cos}\mathrm{2}\theta−\mathrm{2}\right) \\ $$$${now}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{32}}\int_{\mathrm{0}} ^{\pi} {cos}\mathrm{6}\theta+\mathrm{2}{cos}\mathrm{4}\theta−{cos}\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\pi} {d}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{32}}\mid\frac{{sin}\mathrm{6}\theta}{\mathrm{6}}+\frac{\mathrm{2}{sin}\mathrm{4}\theta}{\mathrm{4}}−\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\mid_{\mathrm{0}} ^{\pi} −\frac{\mathrm{1}}{\mathrm{16}}×\pi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{} {\:}\:\underset{{intregal}\:{value}\:{zer}\underset{} {{o}}} {\Downarrow}\:\:\:\:−\frac{\mathrm{1}}{\mathrm{16}}×\pi \\ $$$$ \\ $$$${answer}\:{is}\:=\frac{−\pi}{\mathrm{16}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{sinn}\pi=\mathrm{0}\right] \\ $$$${pls}\:{chek}… \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jan/19
