Question

Integrate \[\log (\sin x + \cos x)dx\] from \[\dfrac{{ - \pi}}{4}\] to \[\dfrac{{\pi}}{4}\].

Answer 1

Hint: In order to determine the integral of the given logarithmic function, \[\log (\sin x + \cos x)dx\]. First, we have to compare the given function with the formula, \[\int\limits_b^a {f(x)dx = \int\limits_o^a {[f(x) + f( - x)]dx} } \]is continuous in interval\[\left[ { - a,a} \right]\], the process of finding integrals is called integration.
and also, the logarithmic formula is \[(\log a + \log b = \log ab)\]used to get the required solution.

Complete step-by-step answer:
We are given integrate the equation\[\log (\sin x + \cos x)dx\] with the interval from \[ - \dfrac{\pi }{4}\] to \[\dfrac{\pi }{4}\].
Now, consider, \[{\rm I} = \int\limits_{\dfrac{{ - \pi }}{4}}^{\dfrac{\pi }{4}} {\log (\sin x + \cos x)dx} \] ----------(1)
Let the function\[f(x)\] is continuous in interval [-a,a], then
\[\int\limits_b^a {f(x)dx = \int\limits_o^a {[f(x) + f( - x)]dx} } \]
\[{\rm I} = \int\limits_0^{\dfrac{\pi }{4}} {\left[ {\log (\sin (x) + \cos (x)) + \log (\sin ( - x) + \cos ( - x))} \right]dx} \]
Expanding the bracket on RHS, since \[\cos ( - x) = \cos x\]but, \[\sin ( - x) = - \sin x\]
\[{\rm I} = \int\limits_0^{\dfrac{\pi }{4}} {\left[ {\log (\sin x + \cos x) + \log (\cos x - \sin x)} \right]dx} \]
Accordingly, we get the logarithmic formula is \[(\log a + \log b = \log ab)\]
\[{\rm I} = \int\limits_0^{\dfrac{\pi }{4}} {\left[ {\log (\sin x + \cos x)(\cos x - \sin x)} \right]dx} \]
We can write this as follows
\[{\rm I} = \int\limits_0^{\dfrac{\pi }{4}} {\log ({{\cos }^2}x - {{\sin }^2}x)dx} \]
 \[{\rm I} = \int\limits_0^{\dfrac{\pi }{4}} {\log (\cos 2x)dx} \] --------(2)
Now, let us consider \[2x = t\]\[ \Rightarrow dx = \dfrac{{dt}}{2}\]
When \[x = 0\]then \[t = 0\]
When \[x = \dfrac{\pi }{4}\] then \[t = 2\left( {\dfrac{\pi }{4}} \right) \Rightarrow t = \dfrac{\pi }{2}\]
Now, we can substitute all the values in the equation (2), then
\[ \Rightarrow {\rm I} = \int\limits_0^t {\log (\cos t)dx} \]
\[ \Rightarrow {\rm I} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\cos t)\left( {\dfrac{{dt}}{2}} \right)} \]
\[ \Rightarrow {\rm I} = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {\log (\cos t)dt} \] --------- (3)
By symmetry we have the\[\log (\sin x)dx = {\rm I} = \log (\cos x)dx\] on the interval of \[\left[ {0,\dfrac{\pi }{2}} \right]\]. So this is true for any even or odd function on this interval. \[\int\limits_0^{\dfrac{\pi }{2}} {\log (\cos x)dx = {\rm I}} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin x)dx} \], as is an exercise from Demidovich problem in analysis. then
\[ \Rightarrow {\rm I} = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {\log (\sin t)dt} \] --------- (4)
Here, adding the equation (3) and (4), we can get
\[ \Rightarrow {\rm I} + {\rm I} = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {\log (\sin t + \cos t)dt} \]
since, \[(\log (a + b) = \log ab)\]
\[ \Rightarrow 2{\rm I} = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {\log (\sin t\cos t)dt} \]
\[ \Rightarrow 2{\rm I} = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {\log \dfrac{{\sin 2t}}{2}dt} \]
Now, we can substitute back ‘t’ value, we can get
\[ \Rightarrow 2{\rm I} = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {(\log (\sin 2x) - \log 2)dx} \]
From the equation (4) \[ \Rightarrow {\rm I} = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {\log (\sin t)dt} \] where, \[t = 2x\], we can get
\[ \Rightarrow 2{\rm I} = {\rm I} - \left( {\dfrac{1}{2}} \right)\dfrac{\pi }{2}\log 2\]
By subtracting ‘\[{\rm I}\]’ on both sides of the equation, now
\[
   \Rightarrow 2{\rm I} - {\rm I} = - \dfrac{\pi }{4}\log 2 \\
   \Rightarrow {\rm I} = - \dfrac{\pi }{4}\log 2 \\
 \]
Therefore, \[{\rm I} = - \dfrac{\pi }{4}\log {e^2}\]
Hence, Integrate\[\log (\sin x + \cos x)dx\]from\[\dfrac{{ - \pi}}{4}\]to\[\dfrac{{\pi}}{4}\] is \[{\rm I} = - \dfrac{\pi }{4}\log {e^2}\]

Note: we note that an exercise from Demidovich problem in analysis we have to remind is\[\int\limits_0^{\dfrac{\pi }{2}} {\log (\cos x)dx = {\rm I}} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin x)dx} \].
In integral, there are two types of integrals in maths are definite integral and Indefinite integral
Definite Integral: An integral that contains the upper and lower limits then it is a definite integral. On a real line, x is restricted to lie. Riemann Integral is the other name of the Definite Integral.A definite Integral is represented as: \[\int\limits_a^b {f(x)dx} \]
Indefinite Integral : Indefinite integrals are defined without upper and lower limits. It is represented as: \[\smallint f(x)dx = F(x) + C\]. Where C is any constant and the function \[f\left( x \right)\]is called the integrand.