Math, asked by vaibhav00001, 1 year ago

integration of (cosx)^9/sinx

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Answers

Answered by Manikumarsingh
1
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


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