Evaluate: \(\displaystyle\int_{-\pi/4}^{\pi/4} \dfrac{1}{1 - \sin x}\,dx\)
Q.12 Evaluate: \(\displaystyle\int_{-\pi/4}^{\pi/4} \dfrac{1}{1 - \sin x}\,dx\)
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Answer: 2
\(\dfrac{1}{1 - \sin x} = \dfrac{1 + \sin x}{1 - \sin^2 x} = \dfrac{1 + \sin x}{\cos^2 x} = \sec^2 x + \sec x\tan x\).
\(I = \left[\tan x + \sec x\right]_{-\pi/4}^{\pi/4} = (1 + \sqrt2) - (-1 + \sqrt2) = 2\).
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· 28 Oct 2026
AI-verified · Oct 2026
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