Assertion (A): int_{frac{pi}{6}}^{frac{pi}{3} | vedclass

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

Question

(A) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} dx}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} \dots (1)$Using the property

Vedclass · Accepted Answer

$A$ is true,$R$ is true and $R$ is the correct explanation of $A$

Vedclass · Answer

(A) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} dx}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} \dots (1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$:$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin(\frac{\pi}{2}-x))^{\sqrt{2}} dx}{(\sin(\frac{\pi}{2}-x))^{\sqrt{2}}+(\cos(\frac{\pi}{2}-x))^{\sqrt{2}}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\cos x)^{\sqrt{2}} dx}{(\cos x)^{\sqrt{2}}+(\sin x)^{\sqrt{2}}} \dots (2)$Adding $(1)$ and $(2)$:$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$$I = \frac{\pi}{12}$. Thus,Assertion $(A)$ is true.For Reason $(R)$,using the property $\int_{a}^{b} \frac{f(x) dx}{f(x)+f(a+b-x)} = \frac{b-a}{2}$:Here $a = \frac{\pi}{6}$ and $b = \frac{\pi}{3}$,so $\frac{b-a}{2} = \frac{\frac{\pi}{3} - \frac{\pi}{6}}{2} = \frac{\frac{\pi}{6}}{2} = \frac{\pi}{12}$.Since the general property holds,Reason $(R)$ is true and correctly explains $(A)$.

Assertion $(A)$: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} dx}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} = \frac{\pi}{12}$
Reason $(R)$: $\int_{a}^{b} \frac{f(x) dx}{f(x)+f(a+b-x)} = \frac{b-a}{2}$

Similar Questions

The value of $\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$ is

If $[\cdot]$ denotes the greatest integer function,then the integral $\int_{0}^{\pi} [\cos x] \, dx$ is equal to:

The value of $\int_0^1 \tan^{-1}(1-x+x^2) dx$ is

$\int_0^{\pi / 2} \log _e(\sin 2 x) d x$

$\int_0^{\pi / 2} \log |\tan x+\cot x| \, dx=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion $(A)$: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} dx}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} = \frac{\pi}{12}$Reason $(R)$: $\int_{a}^{b} \frac{f(x) dx}{f(x)+f(a+b-x)} = \frac{b-a}{2}$