해법
적분 sin(x)cos23(x)
해법
ln∣sin(x)∣−211sin2(x)+455sin4(x)−255sin6(x)+4165sin8(x)−5231sin10(x)+277sin12(x)−7165sin14(x)+16165sin16(x)−1855sin18(x)+2011sin20(x)−22sin22(x)+C
솔루션 단계
∫sin(x)cos23(x)dx
삼각성을 사용하여 다시 쓰기
=∫sin(x)(1−sin2(x))11cos(x)dx
대체 적용
=∫u(1−u2)11du
u(1−u2)11확대한다:u1−11u+55u3−165u5+330u7−462u9+462u11−330u13+165u15−55u17+11u19−u21
=∫u1−11u+55u3−165u5+330u7−462u9+462u11−330u13+165u15−55u17+11u19−u21du
합계 규칙 적용: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=∫u1du−∫11udu+∫55u3du−∫165u5du+∫330u7du−∫462u9du+∫462u11du−∫330u13du+∫165u15du−∫55u17du+∫11u19du−∫u21du
∫u1du=ln∣u∣
∫11udu=211u2
∫55u3du=455u4
∫165u5du=255u6
∫330u7du=4165u8
∫462u9du=5231u10
∫462u11du=277u12
∫330u13du=7165u14
∫165u15du=16165u16
∫55u17du=1855u18
∫11u19du=2011u20
∫u21du=22u22
=ln∣u∣−211u2+455u4−255u6+4165u8−5231u10+277u12−7165u14+16165u16−1855u18+2011u20−22u22
뒤로 대체 u=sin(x)=ln∣sin(x)∣−211sin2(x)+455sin4(x)−255sin6(x)+4165sin8(x)−5231sin10(x)+277sin12(x)−7165sin14(x)+16165sin16(x)−1855sin18(x)+2011sin20(x)−22sin22(x)
솔루션에 상수 추가=ln∣sin(x)∣−211sin2(x)+455sin4(x)−255sin6(x)+4165sin8(x)−5231sin10(x)+277sin12(x)−7165sin14(x)+16165sin16(x)−1855sin18(x)+2011sin20(x)−22sin22(x)+C