intlimits_0^{pi / 2n} frac{dx}{1 + tan^n(nx)} | vedclass

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

Question

(B) माना $I = \int\limits_0^{\pi / 2n} \frac{dx}{1 + 	an^n(nx)}$.$nx = t$ प्रतिस्थापित करने पर,$n \, dx = dt$,अतः $dx = \frac{dt}{n}$.जब $x = 0, t =

Vedclass · Accepted Answer

$\frac{\pi}{4n}$

Vedclass · Answer

(B) माना $I = \int\limits_0^{\pi / 2n} \frac{dx}{1 + 	an^n(nx)}$.$nx = t$ प्रतिस्थापित करने पर,$n \, dx = dt$,अतः $dx = \frac{dt}{n}$.जब $x = 0, t = 0$. जब $x = \frac{\pi}{2n}, t = \frac{\pi}{2}$.$I = \frac{1}{n} \int\limits_0^{\pi / 2} \frac{dt}{1 + 	an^n(t)} = \frac{1}{n} \int\limits_0^{\pi / 2} \frac{\cos^n(t)}{\sin^n(t) + \cos^n(t)} \, dt$.गुणधर्म $\int\limits_0^a f(t) \, dt = \int\limits_0^a f(a-t) \, dt$ का उपयोग करने पर:$I = \frac{1}{n} \int\limits_0^{\pi / 2} \frac{\sin^n(t)}{\cos^n(t) + \sin^n(t)} \, dt$.$I$ के दोनों व्यंजकों को जोड़ने पर:$2I = \frac{1}{n} \int\limits_0^{\pi / 2} \frac{\cos^n(t) + \sin^n(t)}{\cos^n(t) + \sin^n(t)} \, dt = \frac{1}{n} \int\limits_0^{\pi / 2} 1 \, dt = \frac{1}{n} \left[ t ight]_0^{\pi / 2} = \frac{\pi}{2n}$.अतः,$I = \frac{\pi}{4n}$.

$\int\limits_0^{\pi / 2n} \frac{dx}{1 + \tan^n(nx)} = $

Similar Questions

निश्चित समाकलन $\int_{19}^{37} (\{x\}^2 + 3 \sin(2\pi x)) \, dx$ का मान ज्ञात कीजिए,जहाँ $\{x\}$ भिन्नात्मक भाग फलन को दर्शाता है।

$\int_3^5(x-3)^3(5-x)^5 d x=$

$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt[3]{\sec x}}{\sqrt[3]{\sec x}+\sqrt[3]{\operatorname{cosec} x}} d x=$

$\int_2^3 {\frac{{\sqrt x }}{{\sqrt {5 - x} + \sqrt x }}} \,dx$ का मान क्या है?

$\int_{0}^{\pi} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module