Find the integral of log(sinx+cosx) from -pi/4 to pi/4.

Integral is the antiderivative of a function.

Answer: The value of integral of  log (sin x + cos x) from -pi/4 to pi/4 is - π / 4 log 2.

Let's solve this step by step.

Explanation:

As we know that if f (x) is continuous in integral of [- a, a ] then \(\int_{-a}^a \) f (x) dx = \(\int_{-a}^a \) [ f (x) + f (- x) ] dx.

Using the above we will solve I = \(\int_{-\pi/4}^{\pi/4} \) log (sin x + cos x) dx

⇒  \(\int_{-\pi/4}^{\pi/4} \) [ log (sin x + cos x) +  log (sin (- x) + cos (- x) ] dx

⇒  \(\int_{-\pi/4}^{\pi/4} \) [ log (sin x + cos x) + log (cos x - sin x) ] dx

⇒  \(\int_{-\pi/4}^{\pi/4} \)  log (cos² x - sin² x) dx

⇒  \(\int_{-\pi/4}^{\pi/4} \)  log ( cos 2x ) dx

Let 2x = t 

On differentiating both sides we get,

⇒  dx = dt / 2

By substituting the values of t and dt in \(\int_{-\pi/4}^{\pi/4} \) log (cos 2x),

We get x = 0, t = 0 and x = π / 4, t = π / 2

⇒  \(\int_0^{\pi/2} \)  log (cos t ) dt / 2

⇒  1 / 2 \(\int_0^{\pi/2} \)  log (cos t ) dt 

On integrating this we get,

⇒  1 / 2 [ - π / 2 log 2 ]

⇒  - π / 4 log 2 

Thus, the value of integral of  log (sin x + cos x) from -pi/4 to pi/4 is - π / 4 log 2.