# What is the integration of log cos x dx ?

## Solution :

We have, I = $$\int$$ log cos x dx

By using integraton by parts,

I = $$\int$$ 1.log cos x dx

Taking log cos x as first function and 1 as second function. Then,

I = log cos x $$\int$$ 1 dx – $$\int$$ { $${d\over dx}$$ (log cos x) $$\int$$ 1 dx } dx

I = x log cos x – $$\int$$ { $${-sinx\over cos x} x$$ } dx

I = x log cos x + $$\int$$ x tan x dx

Again using integration by parts,

I = x log cos x + x log |sec x | – $$\int$$ {1.(log sec x)} dx

I = x log cos x + x log sec x – I

2I = x log cos x + x log sec x + C

I = $$x log cos x\over 2$$ + $$x log sec x\over 2$$ + C

Hence, the integration of log cos x with respect to x is $$x log cos x\over 2$$ + $$x log sec x\over 2$$ + C

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