int(sqrt{tan x}+sqrt{cot x}) d x= | vedclass

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

Question

(A) ધારો કે $I = \int (\sqrt{	an x} + \sqrt{\cot x}) dx = \int \frac{	an x + 1}{\sqrt{	an x}} dx$. $	an x = t^2$ આદેશ લેતા,$\sec^2 x dx = 2t dt$ મ

Vedclass · Accepted Answer

$\sqrt{2} 	an^{-1}\left(\frac{	an x-1}{\sqrt{2 	an x}}ight)+c$

Vedclass · Answer

(A) ધારો કે $I = \int (\sqrt{	an x} + \sqrt{\cot x}) dx = \int \frac{	an x + 1}{\sqrt{	an x}} dx$. $	an x = t^2$ આદેશ લેતા,$\sec^2 x dx = 2t dt$ મળે. $\sec^2 x = 1 + 	an^2 x = 1 + t^4$ હોવાથી,$dx = \frac{2t}{1+t^4} dt$ થાય. તેથી,$I = \int \frac{t^2 + 1}{t} \cdot \frac{2t}{1+t^4} dt = 2 \int \frac{t^2 + 1}{t^4 + 1} dt$. અંશ અને છેદને $t^2$ વડે ભાગતા: $I = 2 \int \frac{1 + 1/t^2}{t^2 + 1/t^2} dt = 2 \int \frac{1 + 1/t^2}{(t - 1/t)^2 + 2} dt$. $u = t - 1/t$ લેતા,$du = (1 + 1/t^2) dt$ મળે. $I = 2 \int \frac{du}{u^2 + (\sqrt{2})^2} = 2 \cdot \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{u}{\sqrt{2}}ight) + c = \sqrt{2} 	an^{-1}\left(\frac{t - 1/t}{\sqrt{2}}ight) + c$. $t = \sqrt{	an x}$ મૂકતા,$I = \sqrt{2} 	an^{-1}\left(\frac{	an x - 1}{\sqrt{2 	an x}}ight) + c$ મળે.

$\int(\sqrt{\tan x}+\sqrt{\cot x}) d x=$

Similar Questions

$\int \frac{x^{2}+1}{x^{4}+x^{2}+1} dx$ નું મૂલ્ય શું છે?

$\int \sqrt{\frac{a+x}{a-x}} \, dx = $

$\int \frac{\sin x \cdot \sec ^2 x-\tan x \cdot \sin x+\cos x}{(1-\cos 2 x)} d x=$

જો $\int \frac{dx}{2 \cos x + 3 \sin x + 4} = \frac{2}{\sqrt{3}} f(x) + c$ હોય,તો $f\left(\frac{2 \pi}{3}\right) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module