I=∫0π log(1+cos x)dx =∫0π log(1+cos(π−x))dx I=∫0π log(1−cos x)dx 2I=∫0π log(1+cos x)dx (1- cos x)dx2I=∫0πlog Sin2x dx 2I=2 ∫0π log Sin xdx I=2∫0π/2log Sin xdx =2-π2log 2 =-π log 2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\int_{0}^{π} \log (1 + \cos x) d x =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$π \log 2$

$- π \log 2$

$\frac{π}{2} \log 2$

$- \frac{π}{2} \log 2$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$I = \int_{0}^{π} \log (1 + \cos x) dx = \int_{0}^{π} \log (1 + \cos (π - x)) dx I = \int_{0}^{π} \log (1 - \cos x) dx 2 I = \int_{0}^{π} \log (1 + \cos x) dx (1 - \cos x) dx$
$2 I = \int_{0}^{π} \log {Sin}^{2} x dx 2 I = 2 \int_{0}^{π} \log |Sin x| dx I = 2 \int_{0}^{π / 2} \log |Sin x| dx = 2 (\frac{- π}{2} \log 2) = - π \log 2$