求sinα^4(sinα+cosα)^2的不定积分
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求sinα^4(sinα+cosα)^2的不定积分
sinα^4(sinα+cosα)^2
=(1-cos²a)²(1+2sinacosa)
=(1-2cos²a+cos^4a)(1+2sinacosa)
=1+2sinacosa-2cos²a-4sinacos³a+cos^4a+2sinacos^5a
=1+sin(2a)-(cos2a+1)-2sinacos³a+cos^4a+2sinacos^a
=sin(2a)-cos2a-2sinacos³a+cos^4a+2sinacos^5a
∫sin(2a)-cos2a-2sinacos³a+cos^4a+2sinacos^5a da
=-(1/2)cos(2a)-(1/2)sin2a+∫2cos³ad(cosa)+∫cos^4ada-∫2cos^5a d(cosa)
=-(1/2)cos(2a)-(1/2)sin(2a)+(1/2)cos^4a-(1/3)cos^6a+∫cos^4ada
=-(1/2)cos(2a)-(1/2)sin(2a)+(1/2)cos^4a-(1/3)cos^6a+(1/4)cos³asina+(3/4)∫cos²ada
=-(1/2)cos(2a)-(1/2)sin(2a)+(1/2)cos^4a-(1/3)cos^6a+(1/4)cos³asina+(3/4)[(a/2)+(1/4)sin(2a)+C]
=3a/8-(1/2)cos(2a)-(5/16)sin(2a)+(1/2)cos^4a-(1/3)cos^6a+(1/4)cos³asina+C
=(1-cos²a)²(1+2sinacosa)
=(1-2cos²a+cos^4a)(1+2sinacosa)
=1+2sinacosa-2cos²a-4sinacos³a+cos^4a+2sinacos^5a
=1+sin(2a)-(cos2a+1)-2sinacos³a+cos^4a+2sinacos^a
=sin(2a)-cos2a-2sinacos³a+cos^4a+2sinacos^5a
∫sin(2a)-cos2a-2sinacos³a+cos^4a+2sinacos^5a da
=-(1/2)cos(2a)-(1/2)sin2a+∫2cos³ad(cosa)+∫cos^4ada-∫2cos^5a d(cosa)
=-(1/2)cos(2a)-(1/2)sin(2a)+(1/2)cos^4a-(1/3)cos^6a+∫cos^4ada
=-(1/2)cos(2a)-(1/2)sin(2a)+(1/2)cos^4a-(1/3)cos^6a+(1/4)cos³asina+(3/4)∫cos²ada
=-(1/2)cos(2a)-(1/2)sin(2a)+(1/2)cos^4a-(1/3)cos^6a+(1/4)cos³asina+(3/4)[(a/2)+(1/4)sin(2a)+C]
=3a/8-(1/2)cos(2a)-(5/16)sin(2a)+(1/2)cos^4a-(1/3)cos^6a+(1/4)cos³asina+C
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