int_0^{pi / 4} [sqrt{tan x} + sqrt{cot x}] , | vedclass

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

Question

(A) माना $I = \int_0^{\pi / 4} (\sqrt{	an x} + \sqrt{\cot x}) \, dx$. $I = \int_0^{\pi / 4} \left( \frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{

Vedclass · Accepted Answer

$\frac{\pi}{\sqrt{2}}$

Vedclass · Answer

(A) माना $I = \int_0^{\pi / 4} (\sqrt{	an x} + \sqrt{\cot x}) \, dx$. $I = \int_0^{\pi / 4} \left( \frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}} ight) \, dx$. $I = \int_0^{\pi / 4} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$. अंश और हर को $\sqrt{2}$ से गुणा करने पर: $I = \int_0^{\pi / 4} \frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int_0^{\pi / 4} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx$. माना $t = \sin x - \cos x$, तब $dt = (\cos x + \sin x) \, dx$. जब $x = 0$, तब $t = -1$. जब $x = \pi/4$, तब $t = 0$. $I = \sqrt{2} \int_{-1}^0 \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} [\sin^{-1} t]_{-1}^0$. $I = \sqrt{2} [\sin^{-1}(0) - \sin^{-1}(-1)] = \sqrt{2} [0 - (-\pi/2)] = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}}$.

$\int_0^{\pi / 4} [\sqrt{\tan x} + \sqrt{\cot x}] \, dx$ का मान ज्ञात कीजिए।

Similar Questions

$\int_0^1 {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\,dx = } $

यदि $I$, $I_1=\int_0^1 e^{-x} \cos ^2 x \, dx, I_2=\int_0^1 e^{-x^2} \cos ^2 x \, dx, I_3=\int_0^1 e^{-x^2} \, dx, I_4=\int_0^1 e^{-x^2 / 2} \, dx$ में सबसे बड़ा है, तो

$\int_{-2}^{2} |1 - x^2| \, dx = $

$\int_0^{\frac{\pi}{2}} \frac{d x}{\cos x-\sqrt{3} \sin x}=$

$\int_0^2 |2x - 3| \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module