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 $I=\int_{\sqrt{\log _e 2}}^{\sqrt{\log _e 3}} \frac{x \sin x^2}{\sin x^2+\sin \left(\log _e 6-x^2\right)} d x$ is

Evaluate the definite integral: $\int_0^{\pi /2} \log(\tan x) \, dx$.

If $\int\limits_0^1 \frac{\ln x}{\sqrt{1 - x^2}} dx = k \int\limits_0^\pi \ln(1 + \cos x) dx$,then the value of $k$ is:

$\int_{0}^{\pi /2} {\sin 2x \log \tan x \, dx}$ is equal to

The value of $\int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + a^x} dx, a > 0$ is

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 $(R)$: $\int_{a}^{b} \frac{f(x) dx}{f(x)+f(a+b-x)} = \frac{b-a}{2}$