Sarthaks Test
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Evaluate the following Integral:

\(\int\limits_0^1\sqrt{\cfrac{1-\text x}{1+\text x}}d\text x \)

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Let I = \(\int\limits_0^1\sqrt{\cfrac{1-\text x}{1+\text x}}d\text x \)

As we have the trigonometric identity \(\cfrac{1-cos2\theta}{1+cos2\theta}\) = tan2θ, to evaluate this integral we use x = cos 2θ

⇒ dx = –2sin(2θ)dθ (Differentiating both sides)

When x = 0, cos 2θ = 0 ⇒ 2θ =\(\cfrac{\pi}2\) ⇒ θ =\(\cfrac{\pi}4\)

When x = 1, cos 2θ = 1 ⇒ 2θ = 0 ⇒ θ = 0

So, the new limits are \(\cfrac{\pi}4\) and 0.

Substituting this in the original integral,

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