If 0

Question

(C) Let $I = \int \frac{dx}{1-2a \cos x + a^2}$.Using the identity $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$,we have:$I = \int \frac{dx}{1-2a \le

Vedclass · Accepted Answer

$\frac{2}{1-a^2} 	an^{-1}\left[\frac{1+a}{1-a} 	an \frac{x}{2}ight] + c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{1-2a \cos x + a^2}$.Using the identity $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$,we have:$I = \int \frac{dx}{1-2a \left(\frac{1-	an^2(x/2)}{1+	an^2(x/2)}ight) + a^2} = \int \frac{\sec^2(x/2) dx}{(1+a^2)(1+	an^2(x/2)) - 2a(1-	an^2(x/2))}$.Let $t = 	an(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2 dt$.$I = \int \frac{2 dt}{(1+a^2)(1+t^2) - 2a(1-t^2)} = \int \frac{2 dt}{1+a^2+t^2+a^2t^2-2a+2at^2} = \int \frac{2 dt}{(1-a)^2 + t^2(1+a)^2}$.$I = \frac{2}{(1+a)^2} \int \frac{dt}{\left(\frac{1-a}{1+a}ight)^2 + t^2}$.Using $\int \frac{dx}{k^2+x^2} = \frac{1}{k} 	an^{-1}(\frac{x}{k})$,we get:$I = \frac{2}{(1+a)^2} \cdot \frac{1+a}{1-a} 	an^{-1}\left(t \cdot \frac{1+a}{1-a}ight) + c = \frac{2}{1-a^2} 	an^{-1}\left[\frac{1+a}{1-a} 	an \frac{x}{2}ight] + c$.

If $0 < a < 1$,then $\int \frac{dx}{1-2a \cos x + a^2} =$

Similar Questions

If $\int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x=A x+B \log \left(3 e^{2 x}+4\right)+C$,then values of $A$ and $B$ are respectively (where $C$ is a constant of integration.)

$\int \frac{d x}{9 \cos ^2 2 x+16 \sin ^2 2 x}=$

Integrate the function: $\frac{x+2}{\sqrt{4x-x^2}}$

If $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log(f(x)) + c$,then $f(x) = $

If $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$ and $I(0) = 1$,then $I\left(\frac{\pi}{3}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module