Question

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Answer 1

Answer:

∫cotx .log(sinx) dx = {log(sinx)}²/2 + K

Step-by-step explanation:

HII MATE ^_^

We have to integrate , ∫cotx. log(sinx) dx

let log(sinx) = p .......(1)

Differentiating both sides ,

1/sinx × (cosx) = dp/dx

or, cosx/sinx = dp/dx

or, cotx dx = dp ......(2)

Putting the equations (1) and (2) in ∫cotx . log(sinx) dx .

Now, ∫cotx . log(sinx) dx converts into ∫p dp

∫p dp = p²/2 + K   where k is constant.

Now, put p = log(sinx)

Then, ∫cotx .log(sinx) dx = {log(sinx)}²/2 + K

Hence,

∫cotx .log(sinx) dx = {log(sinx)}²/2 + K

HOPE IT HELPS,

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