\int \frac { d x } { \operatorname { cot } \frac { x } { 2 } \cdot \op

1.	\int \frac { d x } { \operatorname { cot } \frac { x } { 2 } \cdot \operatorname { cot } \frac { x } { 3 } \cdot \operatorname { cot } \frac { x } { 6 } }
Answer» 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.

\int \frac { d x } { \operatorname { cot } \frac { x } { 2 } \cdot \operatorname { cot } \frac { x } { 3 } \cdot \operatorname { cot } \frac { x } { 6 } }

Answer»

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.