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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Functions

Evaluate ∫- a...

Question

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

π a

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

π a^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{2} a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

Q. Evaluate:

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

Q. If

y = | x - a | + | x - 2 a | + | x + a |

, then the critical point(s) for the curve is/are

\int_{0}^{\frac{π}{2}} \frac{d x}{a^{2} c o s^{2} x + b^{2} s i n^{2} x}

where (a, b >0) is equal to-

The quadratic polynomial in x whose zeros are -a, 2a is:

Q. The equation of normal to the curve

x = a sin 2 θ, y = a cos θ

θ = \frac{π}{6}

Join BYJU'S Learning Program

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