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

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

Question

(A) ધારો કે $I = \int \frac{dx}{7 + 5\cos x}$.$\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$ સૂત્રનો ઉપયોગ કરતા:$I = \int \frac{dx}{7 + 5 \left( \

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) ધારો કે $I = \int \frac{dx}{7 + 5\cos x}$.$\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$ સૂત્રનો ઉપયોગ કરતા:$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)}$ધારો કે $	an(x/2) = t$,તેથી $\frac{1}{2} \sec^2(x/2) dx = dt$.$I = \int \frac{dt}{t^2 + (\sqrt{6})^2} = \frac{1}{\sqrt{6}} 	an^{-1} \left( \frac{t}{\sqrt{6}} ight) + c$$t = 	an(x/2)$ કિંમત મૂકતા:$I = \frac{1}{\sqrt{6}} 	an^{-1} \left( \frac{	an(x/2)}{\sqrt{6}} ight) + c$.

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

Similar Questions

$x$ ની સાપેક્ષમાં નીચેના વિધેયનું સંકલન કરો: $\frac{\sin(\tan^{-1} x)}{1+x^2}$

જો $\int \frac{(2 x+3)}{x(x+1)(x+2)(x+3)+1} d x =-\frac{1}{p x^2+q x+r}+c$ હોય,તો $\frac{3 p-q}{r}=$

$\int e^{(e^{x}+x)} dx=$

$\int \frac{d x}{\sqrt{2 e^x-1}}=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module