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Mathematics · 2024 · 2 marks
CBSE 2024 · Region 4 · Set 2 · Q25
Evaluate: ∫_(-1/2)^(1/2) cos x · log ((1+x)/(1-x)) d x
Evaluate : $\displaystyle \int_{\frac{-1}{2}}^{\frac{1}{2}} \cos x \cdot \log \left(\frac{1+x}{1-x}\right) \mathrm{d} x$
Marking-scheme solution
$$\text { Let } f(x)=\cos x \cdot \log \left(\frac{1-x}{1+x}\right)
$$
So, $\displaystyle f(-x)=\cos (x) \cdot \log \left(\frac{1+x}{1-x}\right)=-f(x)$ [Odd function]
Thus, $\displaystyle I=\int_{\frac{-1}{2}}^{\frac{1}{2}} \cos x \cdot \log \left(\frac{1-x}{1+x}\right) \mathrm{d} x=0$
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CBSE Class 12 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.