integral root cotx divided by sinx cosx
rohan , 9 Years ago
Grade
Integral Calculus> integral root cotx divided by sinx cosx
1 Answers
KAPIL MANDAL
Last Activity: 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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