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integral root cotx divided by sinx cosx

integral root cotx divided by sinx cosx
               

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1 Answers

KAPIL MANDAL
132 Points
9 years ago
Cotx = cosx/sinx. So,
root cot x /sinx cosx = (cosx)(-1/2) (sinx)(-3/2)                 (1)
Take,
 (cosx)(-1/2)=Z                                           (2)
Differentiating,
(sinx)(-3/2) dx = dZ.                                   (3)
Putting this back in equation (1)
(root cot x /sinx cosx ) dx= (cosx)(-1/2) (sinx)(-3/2) dx        = Z dZ
∫ Z dZ = Z2 / 2   +  C                                   (4)
C is the integration constant.
Putting the value of Z from (2),
the ans is,
(cosx)(-1) /2  +  C
=( 1 / 2cosx)    + C
 

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