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Integrate: (sec^4/9 x)(cosec^14/9 x)

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Solution: Let ,

$I=int sec^4/9xcdot cosec^14/9x hspace0.1cmdx\ \Rightarrow I=int 1/(sinx)^14/9cdot (cosx)^4/9hspace0.1cmdx\ \Rightarrow I=int dx/(sinx/cosx)^14/9cdot (cosx)^4/9cdot (cosx)^14/9\ \Rightarrow I=int sec^2x/(tanx)^14/9hspace0.1cmdx$

put , $tanx=tRightarrow sec^2xdx=dt$

$I=int dt/t^14/9=t^-(5/9)/-(5/9)+c$

$I=-(9/5)(tanx)^-5/9hspace0.1cm+c$

Posted by

Deependra Verma

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