જો k in (1, infty) હોય,તો int frac{1}{1+k cos | vedclass

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

Question

(B) ધારો કે $I = \int \frac{1}{1+k \cos x} d x$. નિત્યસમ $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$ નો ઉપયોગ કરતા: $I = \int \frac{1}{1+k\left(\fr

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) ધારો કે $I = \int \frac{1}{1+k \cos x} d x$. નિત્યસમ $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$ નો ઉપયોગ કરતા: $I = \int \frac{1}{1+k\left(\frac{1-	an^2(x/2)}{1+	an^2(x/2)}ight)} d x = \int \frac{\sec^2(x/2)}{1+	an^2(x/2)+k-k	an^2(x/2)} d x = \int \frac{\sec^2(x/2)}{(k+1)-(k-1)	an^2(x/2)} d x$. $t = 	an(x/2)$ આદેશ લેતા,$dt = \frac{1}{2}\sec^2(x/2) d x$,તેથી $\sec^2(x/2) d x = 2 dt$. $I = 2 \int \frac{dt}{(k+1)-(k-1)t^2} = \frac{2}{k-1} \int \frac{dt}{\left(\sqrt{\frac{k+1}{k-1}}ight)^2 - t^2}$. પ્રમાણિત સંકલન $\int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} ight| + C$ નો ઉપયોગ કરતા: $I = \frac{2}{k-1} \cdot \frac{1}{2 \sqrt{\frac{k+1}{k-1}}} \log \left| \frac{\sqrt{\frac{k+1}{k-1}} + t}{\sqrt{\frac{k+1}{k-1}} - t} ight| + C = \frac{1}{\sqrt{k^2-1}} \log \left| \frac{\sqrt{k+1} + \sqrt{k-1} 	an(x/2)}{\sqrt{k+1} - \sqrt{k-1} 	an(x/2)} ight| + C$.

જો $k \in (1, \infty)$ હોય,તો $\int \frac{1}{1+k \cos x} d x=$

Similar Questions

જો $\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} dx = \left(\frac{23}{41}\right) x + K \log |4 \cos x+5 \sin x| + c$ હોય,તો $K$ ની કિંમત શોધો.

$\int \frac{d x}{\left(2 a x+x^2\right)^{\frac{3}{2}}} = $

નીચેના સંકલિત શોધો:
$\int \frac{1}{1+\tan x} d x$

Vedclass Test Series

Exam Paper Generator

Online Exam Module