If I = intlimits_0^{frac{pi}{2}} ln(sin x) dx | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I_1 = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x + \cos x) dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we

Vedclass · Accepted Answer

$\frac{I}{2}$

Vedclass · Answer

(A) Let $I_1 = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x + \cos x) dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have: $I_1 = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin(-\frac{\pi}{4} + \frac{\pi}{4} - x) + \cos(-\frac{\pi}{4} + \frac{\pi}{4} - x)) dx = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin(-x) + \cos(-x)) dx = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\cos x - \sin x) dx$. Adding the two expressions for $I_1$: $2I_1 = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\ln(\sin x + \cos x) + \ln(\cos x - \sin x)) dx = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\cos^2 x - \sin^2 x) dx = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\cos 2x) dx$. Since $\ln(\cos 2x)$ is an even function,$2I_1 = 2 \int\limits_0^{\frac{\pi}{4}} \ln(\cos 2x) dx$. Let $2x = t$,then $2dx = dt$. When $x=0, t=0$; when $x=\frac{\pi}{4}, t=\frac{\pi}{2}$. $I_1 = \int\limits_0^{\frac{\pi}{2}} \ln(\cos t) \frac{dt}{2} = \frac{1}{2} \int\limits_0^{\frac{\pi}{2}} \ln(\sin t) dt = \frac{I}{2}$.

If $I = \int\limits_0^{\frac{\pi}{2}} \ln(\sin x) dx$,then $\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x + \cos x) dx =$

Similar Questions

$\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$ is equal to :

If $[x]$ is the greatest integer $\leq x$,then the value of the integral $\int_{-0.9}^{0.9} \left( [x^2] + \log \left( \frac{2-x}{2+x} \right) \right) dx$ is

$\int_{-1}^{1} x \tan^{-1} x \, dx$ equals

The value of the integral $\int_{-\pi/4}^{\pi/4} \left( \frac{32 \cos^4 x}{1 + e^{\sin x}} \right) dx$ is:

Let $f$ and $g$ be continuous functions on $[0, a]$ such that $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$,then $\int_0^a f(x) g(x) d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module