Mathematics

# Solve $\int \dfrac{cos^{9}x}{sin x} dx$

##### SOLUTION
$\int { \cfrac { \cos ^{ 9 }{ x } }{ \sin { x } } dx } =\int { \cfrac { \cos ^{ 8 }{ x } \cos { x } }{ \sin { x } } dx } \\ =\int { \cfrac { { (1-\sin ^{ 2 }{ x } ) }^{ 4 }\cos { x } }{ \sin { x } } dx } \quad \quad (\because \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } =1)$
Let $t=\sin { x } \quad \quad dt=\cos { x } dx$
$=\int { \cfrac { { (1-{ t }^{ 2 }) }^{ 4 } }{ t } dt } \\ =\int { \cfrac { { (1+{ t }^{ 4 }-2{ t }^{ 2 }) }^{ 2 } }{ t } dt } \\ =\int { \cfrac { 1+{ t }^{ 8 }+4{ t }^{ 4 }+2{ t }^{ 4 }-4{ t }^{ 2 }-4{ t }^{ 6 } }{ t } dt } \\ =\int { \cfrac { 1 }{ t } dt } +\int { { t }^{ 7 }dt } +4\int { { t }^{ 3 }dt } +2\int { { t }^{ 3 }dt } -4\int { tdt } -4\int { { t }^{ 5 }dt } \\ =\int { \cfrac { 1 }{ t } dt } +\int { { t }^{ 7 }dt } +6\int { { t }^{ 3 }dt } -4\int { tdt } -4\int { { t }^{ 5 }dt } \\ =\ln { \left| t \right| } +\cfrac { { t }^{ 8 } }{ 8 } +\cfrac { 6{ t }^{ 4 } }{ 4 } -\cfrac { 4{ t }^{ 2 } }{ 2 } -\cfrac { 4{ t }^{ 6 } }{ 6 } +C\\ =\ln { \left| t \right| } +\cfrac { { t }^{ 8 } }{ 8 } +\cfrac { 3{ t }^{ 4 } }{ 2 } -2{ t }^{ 2 }-\cfrac { 2 }{ 3 } { t }^{ 6 }+C\\ =\ln { \left| \sin { x } \right| } +\cfrac { \sin ^{ 8 }{ x } }{ 8 } +\cfrac { 3\sin ^{ 4 }{ x } }{ 2 } -2\sin ^{ 2 }{ x } -\cfrac { 2 }{ 3 } \sin ^{ 6 }{ x } +C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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