Question

integration of root cotx/sinxcosx

Answer 1

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