Question

If $\int_{0}^{\pi/2}{}$ log sin x dx = k, then $\int_{0}^{\pi}{}$log (1 + cos x) dx is given by

• A. p log 2 + 4k
• B. p log 2 + 2k
• C. p log 2 + k
• D. p log 9 + k²

Answer 1

Answer: (A)

p log 2 + 4k

Answer 2

I = $\int_{0}^{\pi}{}$log (1 + cos x) dx

= $\int_{0}^{\pi}{}$log (2 cos² $\frac{x}{2}$) dx

= $\int_{0}^{\pi}{}$log 2 dx + $\int_{0}^{\pi}{}$ log cos² $\frac{x}{2}$ dx

= p log 2 + 2$\int_{0}^{\pi}{}$log cos $\frac{x}{2}$ dx

= p log 2 + 4$\int_{0}^{\pi/2}{}$ log cos x dx

= p log 2 + 4 $\int_{0}^{\pi/2}{}$log cos (p/2 – x) dx

= p log 2 + 4 $\int_{0}^{\pi/2}{}$ log sin x dx

= p log 2 + 4k