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प्रश्न
पर्याय
- \[\log\left( \frac{2}{3} \right)\]
- \[\log\left( \frac{3}{2} \right)\]
- \[\log\left( \frac{3}{4} \right)\]
- \[\log\left( \frac{4}{3} \right)\]
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उत्तर
\[Let\, I = \int_0^\frac{\pi}{2} \frac{\cos x}{\left( 2 + \sin x \right)\left( 1 + \sin x \right)} d x\]
\[\text{Let} \sin x , \text{then} \cos x\ dx = dt\]
\[When\ x = 0, t = 0, x = \frac{\pi}{2}, t = 1\]
\[\text{Therefore the integral becomes}\]
\[I = \int_0^1 \frac{dt}{\left( 2 + t \right)\left( 1 + t \right)}\]
\[ = \int_0^1 \left[ \frac{- 1}{2 + t} + \frac{1}{1 + t} \right] dt\]
\[ = \left[ - \log\left( 2 + t \right) + \log\left( 1 + t \right) \right]_0^1 \]
\[ = \left[ \log\left( 1 + t \right) - \log\left( 2 + t \right) \right]_0^1 \]
\[ = \log2 - \log3 - \log1 + \log2\]
\[ = \log\frac{4}{3}\]
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