int_0^{pi / 2} frac{d x}{1+(tan x)^{sqrt{2018 | vedclass

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

Question

(D) माना $I = \int_0^{\pi / 2} \frac{1}{1+(	an x)^{\sqrt{2018}}} dx$  $(i)$गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर:$I = \int

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(D) माना $I = \int_0^{\pi / 2} \frac{1}{1+(	an x)^{\sqrt{2018}}} dx$  $(i)$गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर:$I = \int_0^{\pi / 2} \frac{1}{1+(	an(\pi / 2 - x))^{\sqrt{2018}}} dx$$I = \int_0^{\pi / 2} \frac{1}{1+(\cot x)^{\sqrt{2018}}} dx$$I = \int_0^{\pi / 2} \frac{1}{1+\frac{1}{(	an x)^{\sqrt{2018}}}} dx$$I = \int_0^{\pi / 2} \frac{(	an x)^{\sqrt{2018}}}{1+(	an x)^{\sqrt{2018}}} dx$  $(ii)$समीकरण $(i)$ और $(ii)$ को जोड़ने पर:$2I = \int_0^{\pi / 2} \left( \frac{1}{1+(	an x)^{\sqrt{2018}}} + \frac{(	an x)^{\sqrt{2018}}}{1+(	an x)^{\sqrt{2018}}} ight) dx$$2I = \int_0^{\pi / 2} \frac{1+(	an x)^{\sqrt{2018}}}{1+(	an x)^{\sqrt{2018}}} dx$$2I = \int_0^{\pi / 2} 1 dx$$2I = [x]_0^{\pi / 2} = \frac{\pi}{2}$$I = \frac{\pi}{4}$

$\int_0^{\pi / 2} \frac{d x}{1+(\tan x)^{\sqrt{2018}}}$ का मान किसके बराबर है?

Similar Questions

यदि $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ है,तो:

यदि $I_n = \int_{-\pi}^{\pi} \frac{\sin(nx)}{(1+\pi^x) \sin x} dx$,$n=0, 1, 2, \ldots$,तो
$(A)$ $I_n = I_{n+2}$
$(B)$ $\sum_{m=1}^{10} I_{2m+1} = 10\pi$
$(C)$ $\sum_{m=1}^{10} I_{2m} = 0$
$(D)$ $I_n = I_{n+1}$

यदि $f(x) = \int\limits_0^x {2(\cos^2 3t + 3\sin^2 3t)dt}$ है,तो $f(x + \pi)$ का मान क्या होगा?

$\int_{0}^{1} \frac{dx}{x + \sqrt{1 - x^2}}$ का मान ज्ञात कीजिए।

$\int_{0}^{1} \tan ^{-1}\left[\frac{2 x-1}{1+x-x^{2}}\right] d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module