integarationof dx/cotx/2.cotx/3.cotx/6
Note : ∫ tan ( mx ) dx = ( 1/m )· ln | sec mx | = ( - 1/m )· ln | cos mx | ... ... (1)
We have,
I = ∫ { -1 / [ cot ( x/2 )· cot ( x/3 )· cot ( x/6 ) ] dx
I = ∫ [ - tan(x/2)· tan(x/3)· tan(x/6) ] dx .................... (2)
We have
... tan(x/2 - x/3) = tan(x/6)
⇒ [ tan(x/2) - tan(x/3) ] / [ 1 + tan(x/2)·tan(x/3) ] = tan(x/6)
⇒ tan(x/2) - tan(x/3) = tan(x/6) + tan(x/2)·tan(x/3)·tan(x/6)
⇒ - tan(x/2)·tan(x/3)·tan(x/6) = tan(x/6) + tan(x/3) - tan(x/2) .... (3)
From (1), (2) and (3),
I = ∫ [ tan(x/6) + tan(x/3) - tan(x/2) ] dx
= 6. ln | sec(x/6) | + 3. ln | sec (x/3) | - 2. ln | sec(x/2) | + C ..... Ans.
= 2. ln | cos( x/2 ) | - 3. ln | cos( x/3 ) | - 6. ln | cos( x/6) | + C ... Ans.
We have,
I = ∫ { -1 / [ cot ( x/2 )· cot ( x/3 )· cot ( x/6 ) ] dx
I = ∫ [ - tan(x/2)· tan(x/3)· tan(x/6) ] dx .................... (2)
We have
... tan(x/2 - x/3) = tan(x/6)
⇒ [ tan(x/2) - tan(x/3) ] / [ 1 + tan(x/2)·tan(x/3) ] = tan(x/6)
⇒ tan(x/2) - tan(x/3) = tan(x/6) + tan(x/2)·tan(x/3)·tan(x/6)
⇒ - tan(x/2)·tan(x/3)·tan(x/6) = tan(x/6) + tan(x/3) - tan(x/2) .... (3)
From (1), (2) and (3),
I = ∫ [ tan(x/6) + tan(x/3) - tan(x/2) ] dx
= 6. ln | sec(x/6) | + 3. ln | sec (x/3) | - 2. ln | sec(x/2) | + C ..... Ans.
= 2. ln | cos( x/2 ) | - 3. ln | cos( x/3 ) | - 6. ln | cos( x/6) | + C ... Ans.