IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

Greatdreams

Multiplying numerator and deno by

cosec ² x

we have,

(3 cosec² x + 2 cot x cosec x ) / ( 2 cosec x + 3 cot x ) ²

= - (- 3 cosec ² x- 2 cot x cosec x)/(2 cosec x + 3 cot x ) ²

Now that becomes simple then and we have,

So I = 1/(2 cosec x + 3 cot x) = [sin x/ 2 + 3 cosec x] + c

konichiwa2x

edit: removed solution as per radon's request. Open for others to try

Another method: Substitute

computer001

make given integral as
{(3+2cosx)/3sinx}d(1/2+3cosx)
apply parts
the integral trem comes out as
integarl[ {(2sin^2x+(3+2cosx)cosx)/sin^2x}*dx/(3+2cosx)]
= (3+2cosx/sin^2x)*dx/3+2cosx= cosec^2xdx
which is direct
ofcourse there will b some 1/3 or something outside as 1/3*integral..but thts all not important here

Radon222

wow ! great attempts ....and really awesome substitution by konichiwa

sandeepramesh

yes that method is called integration by differentiation or sth :)

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