The value of int_0^{pi / 2} frac{(cos x)^{sin | vedclass

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

Question

(A) Let $I = \int_0^{\pi / 2} \frac{(\cos x)^{\sin x}}{(\cos x)^{\sin x} + (\sin x)^{\cos x}} dx$  --- $(1)$Using the property $\int_0^a f(x) dx = \in

Vedclass · Accepted Answer

$\pi / 4$

Vedclass · Answer

(A) Let $I = \int_0^{\pi / 2} \frac{(\cos x)^{\sin x}}{(\cos x)^{\sin x} + (\sin x)^{\cos x}} dx$  --- $(1)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we get:$I = \int_0^{\pi / 2} \frac{(\cos(\pi/2 - x))^{\sin(\pi/2 - x)}}{(\cos(\pi/2 - x))^{\sin(\pi/2 - x)} + (\sin(\pi/2 - x))^{\cos(\pi/2 - x)}} dx$$I = \int_0^{\pi / 2} \frac{(\sin x)^{\cos x}}{(\sin x)^{\cos x} + (\cos x)^{\sin x}} dx$  --- $(2)$Adding $(1)$ and $(2)$:$2I = \int_0^{\pi / 2} \frac{(\cos x)^{\sin x} + (\sin x)^{\cos x}}{(\cos x)^{\sin x} + (\sin x)^{\cos x}} dx$$2I = \int_0^{\pi / 2} 1 dx$$2I = [x]_0^{\pi / 2} = \pi / 2$$I = \pi / 4$

The value of $\int_0^{\pi / 2} \frac{(\cos x)^{\sin x}}{(\cos x)^{\sin x}+(\sin x)^{\cos x}} d x$ is

Similar Questions

$\int_0^1 \log \left(\frac{1}{x}-1\right) d x=$

If ${I_n} = \int_{0}^{\pi /4} {\tan^n x} \,dx$,then $\lim_{n \to \infty} n[{I_n} + {I_{n - 2}}]$ equals

$\int_{-1}^{1} \sin^{11} x \, dx$ is equal to

If $\int_0^{\frac{\pi}{2}} \log \cos x \, dx = \frac{\pi}{2} \log \left(\frac{1}{2}\right)$,then $\int_0^{\frac{\pi}{2}} \log \sec x \, dx = $

$\int_0^{\pi / 2} \frac{\sin ^3 x \cos x \, dx}{\sin ^4 x+\cos ^4 x} = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module