The value of I = int_{pi / 2}^{5 pi / 2} frac | vedclass

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

Question

(B) Let $I = \int_{\pi / 2}^{5 \pi / 2} \frac{e^{	an^{-1}(\sin x)}}{e^{	an^{-1}(\sin x)} + e^{	an^{-1}(\cos x)}} dx$. Using the property $\int_{a}^

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(B) Let $I = \int_{\pi / 2}^{5 \pi / 2} \frac{e^{	an^{-1}(\sin x)}}{e^{	an^{-1}(\sin x)} + e^{	an^{-1}(\cos x)}} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$, we have $a+b = \frac{\pi}{2} + \frac{5\pi}{2} = 3\pi$. $I = \int_{\pi / 2}^{5 \pi / 2} \frac{e^{	an^{-1}(\sin(3\pi - x))}}{e^{	an^{-1}(\sin(3\pi - x))} + e^{	an^{-1}(\cos(3\pi - x))}} dx$. Since $\sin(3\pi - x) = \sin x$ and $\cos(3\pi - x) = -\cos x$, this does not simplify directly. Let us use the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ on the interval $[0, 2\pi]$. The integrand $f(x) = \frac{e^{	an^{-1}(\sin x)}}{e^{	an^{-1}(\sin x)} + e^{	an^{-1}(\cos x)}}$ has a period of $\frac{\pi}{2}$. $I = \int_{\pi/2}^{5\pi/2} f(x) dx = \int_{0}^{2\pi} f(x) dx$. Since $f(x)$ is periodic with period $\frac{\pi}{2}$, $\int_{0}^{2\pi} f(x) dx = 4 \int_{0}^{\pi/2} f(x) dx$. Let $J = \int_{0}^{\pi/2} \frac{e^{	an^{-1}(\sin x)}}{e^{	an^{-1}(\sin x)} + e^{	an^{-1}(\cos x)}} dx$. Using $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$, $J = \int_{0}^{\pi/2} \frac{e^{	an^{-1}(\cos x)}}{e^{	an^{-1}(\cos x)} + e^{	an^{-1}(\sin x)}} dx$. $2J = \int_{0}^{\pi/2} 1 dx = \frac{\pi}{2} \implies J = \frac{\pi}{4}$. Thus, $I = 4 	imes \frac{\pi}{4} = \pi$.

The value of $I = \int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)} + e^{\tan^{-1}(\cos x)}} dx$ is

Similar Questions

Let $I = \int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} dx$. Then

$\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} d x=$

The integral $\int_{\frac{-1}{2}}^{\frac{1}{2}} \left([x] + \log_{e}\left(\frac{1+x}{1-x}\right)\right) dx$,where $[x]$ represents the greatest integer function,equals:

If $a > 0$, then $\int_{-\pi}^{\pi} \frac{\sin^2 x}{1+a^x} dx$ is equal to

If $f(x) = \frac{e^x}{1+e^x}$, $l_1 = \int_{f(-a)}^{f(a)} x g(x(1-x)) dx$ and $l_2 = \int_{f(-a)}^{f(a)} g(x(1-x)) dx$, then the value of $\frac{l_2}{l_1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module