Evaluate the integral: int frac{dx}{2+cos x} | vedclass

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

Question

(A) To evaluate the integral $I = \int \frac{dx}{2+\cos x}$,we use the substitution $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$.Substituting this i

Vedclass · Accepted Answer

$\frac{2}{\sqrt{3}} 	an^{-1}\left(\frac{1}{\sqrt{3}} 	an \frac{x}{2}ight) + c$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{dx}{2+\cos x}$,we use the substitution $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$.Substituting this into the integral,we get:$I = \int \frac{dx}{2 + \frac{1-	an^2(x/2)}{1+	an^2(x/2)}} = \int \frac{(1+	an^2(x/2)) dx}{2(1+	an^2(x/2)) + 1 - 	an^2(x/2)} = \int \frac{\sec^2(x/2) dx}{3 + 	an^2(x/2)}$.Let $t = 	an(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx$,which implies $\sec^2(x/2) dx = 2 dt$.Substituting these into the integral:$I = \int \frac{2 dt}{3 + t^2} = 2 \int \frac{dt}{(\sqrt{3})^2 + t^2}$.Using the standard integral formula $\int \frac{dx}{a^2+x^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$,we get:$I = 2 \cdot \frac{1}{\sqrt{3}} 	an^{-1}\left(\frac{t}{\sqrt{3}}ight) + c = \frac{2}{\sqrt{3}} 	an^{-1}\left(\frac{1}{\sqrt{3}} 	an \frac{x}{2}ight) + c$.

Evaluate the integral: $\int \frac{dx}{2+\cos x}$

Similar Questions

If $\int \frac{\log (1+x^4)}{x^3} d x=f(x) \log \left(\frac{1}{g(x)}\right)+\tan ^{-1}(h(x))+c$,then $h(x)\left[f(x)+f\left(\frac{1}{x}\right)\right]=$

For any integer $n \geq 2$,if $I_n = \int \cot^n x \, dx$,then $I_5 =$

$\int \frac{dx}{(x+1) \sqrt{x^2+1}} = $

If $\int \frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} dx = Ax + B \log |9e^{2x} - 4| + C$,then $A, B$ and $C$ are

The value of $\int e^{\tan ^{-1} x} \cdot \frac{1+x+x^2}{1+x^2} dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module