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CBSE Class 12 Math 2008 Solved Paper

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Question : 26
Total: 29
Evaluate:
a
∫
−a
√
a−x
a+x
 dx
Solution:  
I =
a
∫
−a
√
a−x
a+x
 dx
=
a
∫
−a
a−x
√a2−x2
 dx
a
∫
−a
a
√a2−x2
dx -
a
∫
−a
x
√a2−x2
 dx
= I1+I2
Where I1 =
a
∫
−a
a
√a2−x2
 dx , which is the integral of an even function
And I2 =
a
∫
−a
x
√a2−x2
 , which is the integral of an odd function, and so I2 = 0
Now, I = I1 =
a
∫
−a
a
√a2−x2
 dx
= 2
a
∫
0
a
√a2−x2
 dx
2a
a
∫
0
1
√a2−x2
 dx
= 2a |sin−1(
x
a
)|0a
= 2a |sin−11−sin−10|
= 2a (
π
2
)
= πa
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