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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