Solve the equaton

**zorro** · May 3rd 2009, 10:41 PM

$ \int\limits_{-\pi/2}^{\pi/2} sin^2x cos^2x (sinx+cosx)dx. $

**skeeter** · May 4th 2009, 05:36 AM

Originally Posted by zorro

$ \int\limits_{-\pi/2}^{\pi/2} sin^2x cos^2x (sinx+cosx)dx. $

$\sin^2{x} \cos^2{x} (\sin{x}+\cos{x}) =$

$\sin^2{x} \cos^2{x} \sin{x} + \sin^2{x} \cos^2{x} \cos{x} = $

$(1 - \cos^2{x}) \cos^2{x} \sin{x} + \sin^2{x}(1 - \sin^2{x})\cos{x} =$

$(\cos^2{x} - \cos^4{x})\sin{x} + (\sin^2{x} - \sin^4{x})\cos{x}$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^4{x} - \cos^2{x})(-\sin{x}) \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\sin^2{x} - \sin^4{x})\cos{x} \, dx$

for the first integral, let $u = \cos{x}$

for the second integral, let $u = \sin{x}$

**zorro** · December 27th 2009, 12:22 PM

Originally Posted by skeeter

$\sin^2{x} \cos^2{x} (\sin{x}+\cos{x}) =$

$\sin^2{x} \cos^2{x} \sin{x} + \sin^2{x} \cos^2{x} \cos{x} = $

$(1 - \cos^2{x}) \cos^2{x} \sin{x} + \sin^2{x}(1 - \sin^2{x})\cos{x} =$

$(\cos^2{x} - \cos^4{x})\sin{x} + (\sin^2{x} - \sin^4{x})\cos{x}$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^4{x} - \cos^2{x})(-\sin{x}) \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\sin^2{x} - \sin^4{x})\cos{x} \, dx$

for the first integral, let $u = \cos{x}$

for the second integral, let $u = \sin{x}$

how did u get a $- sinx$ ??????

**zorro** · December 27th 2009, 12:32 PM

Originally Posted by skeeter

$\sin^2{x} \cos^2{x} (\sin{x}+\cos{x}) =$

$\sin^2{x} \cos^2{x} \sin{x} + \sin^2{x} \cos^2{x} \cos{x} = $

$(1 - \cos^2{x}) \cos^2{x} \sin{x} + \sin^2{x}(1 - \sin^2{x})\cos{x} =$

$(\cos^2{x} - \cos^4{x})\sin{x} + (\sin^2{x} - \sin^4{x})\cos{x}$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^4{x} - \cos^2{x})(-\sin{x}) \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\sin^2{x} - \sin^4{x})\cos{x} \, dx$

for the first integral, let $u = \cos{x}$

for the second integral, let $u = \sin{x}$

$I_1 = \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} (cos^4 x - cos^2 x)(- sinx)dx$

u = cosx ; du = - sinxdx

$= \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} (u^4 - u^2 )du$

$\frac{1}{5} \left[ cos \left( - \frac{\pi}{2} \right) - cos \left( \frac{ \pi}{2} \right) \right]^5$ $- \frac{1}{3} \left[ cos \left( - \frac{\pi}{2} \right) - cos \left( \frac{ \pi}{2} \right) \right]^3$

= 0 ...........Is this correct ?????

Similarly

$I_2 = \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} (sin^2 x - sin^4 x)(cosx)dx$

v = sinx ; dv = cosxdx

$= - \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} (u^2 - u^4 )du$

$- \frac{1}{3} \left[ sin \left( - \frac{\pi}{2} \right) - sin \left( \frac{ \pi}{2} \right) \right]^3$ $- \frac{1}{5} \left[ sin \left( - \frac{\pi}{2} \right) - sin \left( \frac{ \pi}{2} \right) \right]^5$

= $\frac{1}{3} (-1-1) - \frac{1}{5} (-1-1)$

= $-\frac{2}{3} + \frac{2}{5}$

= $- \frac{4}{15}$ ...............Is this correct????

**skeeter** · December 27th 2009, 12:33 PM

Originally Posted by zorro

how did u get a $- sinx$ ??????

digging up the old posts, aren't you?

anyway ...

$(\cos^2{x} - \cos^4{x})\sin{x} = (\cos^4{x} - \cos^2{x})(-\sin{x}) $

**zorro** · December 27th 2009, 11:28 PM

Originally Posted by skeeter

digging up the old posts, aren't you?

anyway ...

$(\cos^2{x} - \cos^4{x})\sin{x} = (\cos^4{x} - \cos^2{x})(-\sin{x}) $

mite but is my work correct ????

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