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प्रश्न
If `int_0^h1/(2+8x^2)dx=pi/16 `then find the value of h.
उत्तर
`int_0^h "dx"/(2 + "8x"^2) = pi/16`
`1/8 int_0^h "dx"/((1/2)^2 + "x"^2) = pi/16`
`int_0^"h" "dx"/("x"^2 + "h"^2) = pi/2`
`2 "tan"^-1 (2"x")]_0^"h" = pi/2`
`2 "tan"^-1 2 "h" = pi/2`
`"tan"^-1 "2h" = pi/4`
2h = 1
h = `1/2`
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