I am trying to find this by using integration by parts but I am not sure how to do it.∫0π2(sin⁡x)7(cos⁡x)5dx

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nastaja1en · Accepted Answer

To integrate the functionf(x)=sinnx⋅cosmx,when n or m are positive odd numbers, we can apply a general technique which consists of expanding f(x) into a sum of terms of the formsinpx⋅cos⁡x, p=1,2,…orcosqx⋅sin⁡x, q=1,2,…Using the identitycos2x=1−sin2xin the formcos4x=(1−sin2x)2=1−2sin2x+sin4x,we rewrite ourf(x)=sin7x⋅cos5x=sin7x⋅cos4x⋅cos⁡x asasf(x)=sin7x⋅(1−2sin2x+sin4x)⋅cos⁡x=sin7x⋅cos⁡x−2sin9x⋅cos⁡x+sin11x⋅cos⁡xEach term is of the form sinpx⋅cos⁡x and can easily be integrated by the substitution u=sin⁡x, u′=cos⁡x, du=cos⁡xdx=u′dx:∫sinpx⋅cos⁡xdx=∫updu=up+1p+1=sinp+1xp+1+C∫0π2sinpx⋅cos⁡xdx=1p+1∫0π2f(x)dx=∫0π2sin7xcos5xdx

shotokan0758s · Accepted Answer

Let I=∫0π2(sin⁡x)7(cos⁡x)5dxThen using the substitution u=π2−x we have I=∫0π2(sin⁡x)5(cos⁡x)7dxAdding gives2I=∫0π2(sin⁡x)5(cos⁡x)5dxSince cos2(x)+sin2(x)=1And that2I=125∫0π2(sin⁡2x)5dxsince sin⁡(2x)=2sin⁡(x)cos⁡(x). Then substituting u=2x gives2I=126∫0π(sin⁡u)5du=126∫0π(sin⁡u)(1−cos2u)2du=126∫0π(sin⁡u)(1−2cos2u+cos4(u))duHenceI=127[−cos⁡u+2cos3u3−cos5u5]0πI=127[2+−2⋅23+25]0πHenceI=123⋅15=1120

I am trying to find this by using integration by

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