29. cot x log sin x

Open in App

Solution

I= ∫ cotxlogsinxdx ( 1 )

Consider logsinx=t

Differentiating both sides,

1 sinx ×cosxdx=dt cotxdx=dt

Substitute in equation ( 1 )

I= ∫ cotxlogsinxdx I= ∫ tdt I=[ t 1+1 1+1 ]+C I= t 2 2 +C I= 1 2 t 2 +C ( 2 )

Substitute t=logsinx in equation ( 2 )

I= 1 2 ( logsinx ) 2 +C

Thus, the integral is I= 1 2 ( logsinx ) 2 +C

Suggest Corrections

Similar questions

cot x log sin x

cot x log sin x

lim x \to 0 x log sin x

The unit vector in the direction of the resultant of vectors $\to a = 2^i + 2^j - 5^k a n d \to b = 2^i +^j + 3^k$ is

Join BYJU'S Learning Program