Question

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

Answer 1

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.