int x operatorname{Tan}^{-1} sqrt{frac{1+x^2} | vedclass

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

Question

(B) ધારો કે $I = \int x \operatorname{Tan}^{-1} \sqrt{\frac{1+x^2}{1-x^2}} \, dx$. $x^2 = \cos 	heta$ લેતા,$2x \, dx = -\sin 	heta \, d	heta$,એટલે

Vedclass · Accepted Answer

$\frac{x^2}{4}\left(\pi-\operatorname{Cos}^{-1} x^2\right)+\frac{1}{4} \sqrt{1-x^4}+c$

Vedclass · Answer

(B) ધારો કે $I = \int x \operatorname{Tan}^{-1} \sqrt{\frac{1+x^2}{1-x^2}} \, dx$. $x^2 = \cos 	heta$ લેતા,$2x \, dx = -\sin 	heta \, d	heta$,એટલે કે $x \, dx = -\frac{1}{2} \sin 	heta \, d	heta$. સંકલન $I = \int \operatorname{Tan}^{-1} \sqrt{\frac{1+\cos 	heta}{1-\cos 	heta}} \left(-\frac{1}{2} \sin 	hetaight) d	heta$ બને છે. $\sqrt{\frac{1+\cos 	heta}{1-\cos 	heta}} = \cot(	heta/2) = 	an(\frac{\pi}{2} - \frac{	heta}{2})$ નો ઉપયોગ કરતા. તેથી,$I = -\frac{1}{2} \int (\frac{\pi}{2} - \frac{	heta}{2}) \sin 	heta \, d	heta = -\frac{\pi}{4} \int \sin 	heta \, d	heta + \frac{1}{4} \int 	heta \sin 	heta \, d	heta$. $I = \frac{\pi}{4} \cos 	heta + \frac{1}{4} [-	heta \cos 	heta + \int \cos 	heta \, d	heta] = \frac{\pi}{4} \cos 	heta - \frac{1}{4} 	heta \cos 	heta + \frac{1}{4} \sin 	heta + c$. $\cos 	heta = x^2$,$	heta = \operatorname{Cos}^{-1} x^2$,અને $\sin 	heta = \sqrt{1-x^4}$ હોવાથી,$I = \frac{x^2}{4} (\pi - \operatorname{Cos}^{-1} x^2) + \frac{1}{4} \sqrt{1-x^4} + c$.

$\int x \operatorname{Tan}^{-1} \sqrt{\frac{1+x^2}{1-x^2}} \, dx =$

Similar Questions

$\int \frac{1}{\cos x \sqrt{\cos 2 x}} \, dx = $ (જ્યાં $C$ એ સંકલનનો અચળાંક છે.)

જો $\int \log \left(6 \sin ^2 x+17 \sin x+12\right)^{\cos x} d x=f(x)+c$ હોય,તો $f\left(\frac{\pi}{2}\right)=$

જો $\int {f(x)\,dx = f(x)} ,$ હોય,તો $\int {{{\left[ {f(x)} \right]}^2}\,dx}$ શું થાય?

$\int \frac{dx}{x[(\log x)^2 + 4\log x - 1]} = $

$\int \frac{\sec x}{\sqrt{\log (\sec x+\tan x)}} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module