integration of (cosx)^9/sinx
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Let I=∫cos9x/sinxdx
=∫cos8xcosx/sinxdx
=∫(cos2x)4cosx/sinxdx
=∫(1−sin2x)4cosx/sinxdx
Let sinx=u
⟹du=cosxdx
⟹I=∫(1−u2)4/udu
=∫1−4u2+6u4−4u6+u8/udu
=∫(1/u−4u+6u3−4u5+u7)du
=ln|u|−2u2+3/2u4−2/3u6+u8/8+C
=ln|sinx|−2sin2x+3/2sin4x−2/3sin6x+sin8x/8+C
=∫cos8xcosx/sinxdx
=∫(cos2x)4cosx/sinxdx
=∫(1−sin2x)4cosx/sinxdx
Let sinx=u
⟹du=cosxdx
⟹I=∫(1−u2)4/udu
=∫1−4u2+6u4−4u6+u8/udu
=∫(1/u−4u+6u3−4u5+u7)du
=ln|u|−2u2+3/2u4−2/3u6+u8/8+C
=ln|sinx|−2sin2x+3/2sin4x−2/3sin6x+sin8x/8+C
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