Question:
\(\int \dfrac{1}{7 + 5\cos x}\,dx =\)
\(\int \dfrac{1}{7 + 5\cos x}\,dx\) का मान क्या है?
Options:
- \(\dfrac{1}{7}\tan^{-1}\left(\tan\dfrac{x}{2}\right) + c\)
- \(\dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{1}{\sqrt{6}}\tan\dfrac{x}{2}\right) + C\)
- \(\dfrac{1}{\sqrt{3}}\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\tan\dfrac{x}{2}\right) + C\)
- \(\dfrac{1}{4}\tan^{-1}\left(\tan\dfrac{x}{2}\right) + C\)
- \(\dfrac{1}{7}\tan^{-1}\left(\tan\dfrac{x}{2}\right) + c\)
- \(\dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{1}{\sqrt{6}}\tan\dfrac{x}{2}\right) + C\)
- \(\dfrac{1}{\sqrt{3}}\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\tan\dfrac{x}{2}\right) + C\)
- \(\dfrac{1}{4}\tan^{-1}\left(\tan\dfrac{x}{2}\right) + C\)
Correct Answer: b)
Explanation:
Substitute \(t = \tan(x/2) \implies dx = \dfrac{2dt}{1+t^2}\) and \(\cos x = \dfrac{1-t^2}{1+t^2}\):
\(\int \dfrac{2dt}{7(1+t^2) + 5(1-t^2)} = \int \dfrac{2dt}{12 + 2t^2} = \int \dfrac{dt}{6 + t^2} = \dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{t}{\sqrt{6}}\right) + C = \dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{1}{\sqrt{6}}\tan\dfrac{x}{2}\right) + C\).
\(t = \tan(x/2)\) प्रतिस्थापित करने पर:
\(\int \dfrac{dt}{6+t^2} = \dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{1}{\sqrt{6}}\tan\dfrac{x}{2}\right) + C\)।