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If int0^k1/(2+8x^2)dx=pi/(16),find the ...

If `int_0^k1/(2+8x^2)dx=pi/(16)`,find the value of `2k`

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The correct Answer is:
1
Givn, `int_0^k1/(2+8x^2)dx=pi/(16)`
= `1/8int_0^k1/((1/2)+x^2)dx=pi/(16)`
= `1/8[2tan^ (-1)2x]_0^kdx=pi/(16)`
= `1/8[tan^ (-1)2k-tan^ (-1)0]_0^kdx=pi/(16)`
...
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