Solution :
We have, I = ∫ log cos x dx
By using integraton by parts,
I = ∫ 1.log cos x dx
Taking log cos x as first function and 1 as second function. Then,
I = log cos x ∫ 1 dx – ∫ { ddx (log cos x) ∫ 1 dx } dx
I = x log cos x – ∫ { −sinxcosxx } dx
I = x log cos x + ∫ x tan x dx
Again using integration by parts,
I = x log cos x + x log |sec x | – ∫ {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 = xlogcosx2 + xlogsecx2 + C
Hence, the integration of log cos x with respect to x is xlogcosx2 + xlogsecx2 + C
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