integrate 1/(sin^6x+cos^6x)
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sin6x+cos6x=(sin2x)3+(cos2x)3
as a3+b3=(a+b)(a2−ab+b2),
so=(sin2x+cos2x)
(sin4x+cos4−cos2x sin2x)=((sin2+cos2x)2−2sin2x cos2x−sin2x cos2x)=(1−3sin2x cos2x)=(1−34sin22x)=14(4−3sin22x)I=∫dxsin6x+cos6x=14∫dx4−3sin22x=14∫−13sin22x−4dx=−14∫dx3sin22x−4dxTake u=2xdu=2 dxSo I=−18∫13sin2u−4du=−18∫sec2u(−1tan2u+4)duTake v=tanudv=sec2u duSo, I=18∫1v2+4dvTake w=v2dw=dv2So I=116∫1w2+1dw=116tan−1w+C=116tan−1(v2)+C=116tan−1(tanu2)+C=116tan−1(tan2x2)+C
as a3+b3=(a+b)(a2−ab+b2),
so=(sin2x+cos2x)
(sin4x+cos4−cos2x sin2x)=((sin2+cos2x)2−2sin2x cos2x−sin2x cos2x)=(1−3sin2x cos2x)=(1−34sin22x)=14(4−3sin22x)I=∫dxsin6x+cos6x=14∫dx4−3sin22x=14∫−13sin22x−4dx=−14∫dx3sin22x−4dxTake u=2xdu=2 dxSo I=−18∫13sin2u−4du=−18∫sec2u(−1tan2u+4)duTake v=tanudv=sec2u duSo, I=18∫1v2+4dvTake w=v2dw=dv2So I=116∫1w2+1dw=116tan−1w+C=116tan−1(v2)+C=116tan−1(tanu2)+C=116tan−1(tan2x2)+C
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