Evaluate - a a - a - x - a - x dxA- TaB- - a2C- -2 aD- 2 a

The correct option is A π a∫^a_ aa/√(a^2 x^2)dx ∫^a_ ax/√(a^2 x^2)dx =2a∫^a_0 1/√(a^2 x^2)dx 0 [∵x/√(a^2 x^2) is odd function ] =2a sin^ 1x/a|^a_0 = 2a ...

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Byju's Answer

Standard XII

Mathematics

Functions

Evaluate ∫- a...

Question

Evaluate $\int_{- a}^{a} \sqrt{\frac{a - x}{a + x}} d x$

π a

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π a^{2}

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2 a

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\frac{π}{2} a

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Solution

The correct option is A $π a$
$\int_{- a}^{a} \frac{a}{\sqrt{a^{2} - x^{2}}} d x - \int_{- a}^{a} \frac{x}{\sqrt{a^{2} - x^{2}}} d x$
$= 2 a \int_{0}^{a} \frac{1}{\sqrt{a^{2} - x^{2}}} d x - 0 [∵ \frac{x}{\sqrt{a^{2} - x^{2}}} i s o d d f u n c t i o n]$
$= 2 a s i n^{- 1} \frac{x}{a} |_{0}^{a} = 2 a [s i n^{- 1} (1) - s i n^{- 1} (0)] = π a$

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Evaluate ∫a−a√a−xa+xdx

The correct option is A πa∫a−aa√a2−x2dx−∫a−ax√a2−x2dx =2a∫a01√a2−x2dx−0[∵x√a2−x2is odd function] =2a sin−1xa|a0=2a[sin−1(1)−sin−1(0)]=πa

Evaluate $\int_{- a}^{a} \sqrt{\frac{a - x}{a + x}} d x$