intlimits_{frac{1}{2}}^2 {frac{1}{x}{{tan }^{ | vedclass

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

Question

(C) ધારો કે $I = \int\limits_{\frac{1}{2}}^2 {\frac{1}{x}{{	an }^{2015}}\left( {x - \frac{1}{x}} ight)dx} $. $x = \frac{1}{t}$ આદેશ લેતા,$dx = -\fr

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) ધારો કે $I = \int\limits_{\frac{1}{2}}^2 {\frac{1}{x}{{	an }^{2015}}\left( {x - \frac{1}{x}} ight)dx} $. $x = \frac{1}{t}$ આદેશ લેતા,$dx = -\frac{1}{t^2} dt$ મળે. જ્યારે $x = \frac{1}{2}$,ત્યારે $t = 2$ અને જ્યારે $x = 2$,ત્યારે $t = \frac{1}{2}$. તેથી,$I = \int\limits_2^{\frac{1}{2}} {t \cdot {{	an }^{2015}}\left( {\frac{1}{t} - t} ight) \cdot \left( -\frac{1}{t^2} ight) dt} $. $I = \int\limits_{\frac{1}{2}}^2 {\frac{1}{t} \cdot {{	an }^{2015}}\left( {-(t - \frac{1}{t})} ight) dt} $. આપણે જાણીએ છીએ કે $	an(- 	heta) = -	an(	heta)$,તેથી $	an^{2015}(-(t - \frac{1}{t})) = -	an^{2015}(t - \frac{1}{t})$. આમ,$I = - \int\limits_{\frac{1}{2}}^2 {\frac{1}{t} {{	an }^{2015}}\left( {t - \frac{1}{t}} ight) dt} = -I$. $2I = 0 \Rightarrow I = 0$.

$\int\limits_{\frac{1}{2}}^2 {\frac{1}{x}{{\tan }^{2015}}\left( {x - \frac{1}{x}} \right)dx} $ ની કિંમત શોધો.

Similar Questions

$\int_0^1 \frac{\log x}{\sqrt{1 - x^2}} \, dx = $

$\int_{-a}^{a} \sin x \, f(\cos x) \, dx = $

$\int_{\pi /4}^{3\pi /4} \frac{\phi}{1 + \sin \phi} \, d\phi$ નું મૂલ્ય શું છે?

$\int_{5 \pi}^{25 \pi}|\sin 2 x+\cos 2 x| d x=$ ($\sqrt{2}$ માં)

$\int_{0}^{\pi} \frac{x \, dx}{1+\cos \alpha \sin x}, (0 < \alpha < \pi)$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module