int frac{dx}{7 + 5cos x} = | vedclass

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

Question

(A) Let $I = \int \frac{dx}{7 + 5\cos x}$.Using the substitution $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$,we get:$I = \int \frac{dx}{7 + 5 \

Vedclass · Accepted Answer

$\frac{1}{\sqrt{6}} 	an^{-1} \left( \frac{1}{\sqrt{6}} 	an \frac{x}{2} ight) + c$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{7 + 5\cos x}$.Using the substitution $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$,we get:$I = \int \frac{dx}{7 + 5 \left( \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)} ight)} = \int \frac{(1 + 	an^2(x/2)) dx}{7(1 + 	an^2(x/2)) + 5(1 - 	an^2(x/2))}$$I = \int \frac{\sec^2(x/2) dx}{7 + 7	an^2(x/2) + 5 - 5	an^2(x/2)} = \int \frac{\sec^2(x/2) dx}{12 + 2	an^2(x/2)}$$I = \frac{1}{2} \int \frac{\sec^2(x/2) dx}{6 + 	an^2(x/2)}$Let $	an(x/2) = t$,then $\frac{1}{2} \sec^2(x/2) dx = dt$.$I = \int \frac{dt}{t^2 + (\sqrt{6})^2} = \frac{1}{\sqrt{6}} 	an^{-1} \left( \frac{t}{\sqrt{6}} ight) + c$Substituting back $t = 	an(x/2)$:$I = \frac{1}{\sqrt{6}} 	an^{-1} \left( \frac{	an(x/2)}{\sqrt{6}} ight) + c$.

$\int \frac{dx}{7 + 5\cos x} = $

Similar Questions

$\int \frac{dx}{1 + 3\sin^2 x} = $

$\int \frac{1}{x(\log x)^2} \, dx = $

$\int \frac{x e^{\left(\frac{x^2}{x^2-2}\right)}}{x^4-4 x^2+4} d x=$

$\int \frac{3 x^9+7 x^8}{\left(x^2+2 x+5 x^8\right)^2} d x=$

$\int \frac{(\tan^{-1} x)^3}{1 + x^2} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module