int frac{dx}{7+5 cos x} ની કિંમત શોધો. | vedclass

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(B) ધારો કે $I = \int \frac{dx}{7+5 \cos x}$.નિત્યસમ $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ અને $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) ધારો કે $I = \int \frac{dx}{7+5 \cos x}$.નિત્યસમ $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ અને $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$ નો ઉપયોગ કરતા:$I = \int \frac{dx}{7(\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}) + 5(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2})}$$I = \int \frac{dx}{12 \cos^2 \frac{x}{2} + 2 \sin^2 \frac{x}{2}}$અંશ અને છેદને $\cos^2 \frac{x}{2}$ વડે ભાગતા:$I = \int \frac{\sec^2 \frac{x}{2} dx}{12 + 2 	an^2 \frac{x}{2}} = \frac{1}{2} \int \frac{\sec^2 \frac{x}{2} dx}{6 + 	an^2 \frac{x}{2}}$ધારો કે $	an \frac{x}{2} = z$,તેથી $\frac{1}{2} \sec^2 \frac{x}{2} dx = dz$.$I = \int \frac{dz}{6 + z^2} = \int \frac{dz}{(\sqrt{6})^2 + z^2}$સૂત્ર $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1} \frac{x}{a} + c$ નો ઉપયોગ કરતા:$I = \frac{1}{\sqrt{6}} 	an^{-1} \frac{z}{\sqrt{6}} + c = \frac{1}{\sqrt{6}} 	an^{-1} \left(\frac{1}{\sqrt{6}} 	an \frac{x}{2}ight) + c$.

$\int \frac{dx}{7+5 \cos x}$ ની કિંમત શોધો.

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