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If I_1 = int_(-π/4)^(π/4) (dx)/(1 + cos 2x) and I_2 = int_(-1/2)^(1/2) |x|dx, then show that I1 − 4I2 = 0. - Mathematics

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Question

If `I_1 = int_(-π/4)^(π/4) (dx)/(1 + cos 2x) and I_2 = int_(-1/2)^(1/2) |x|dx`, then show that I1 − 4I2 = 0.

Sum
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Solution

`I_1 = int_(-π/4)^(π/4) (dx)/(1 + cos 2x)`

Here, using:

1 + cos 2x = 2 cos2x

`I_1 = int_(-π/4)^(π/4) (dx)/(2 cos^2x)`

`I_1 = 1/2 int_(-π/4)^(π/4) sec^2x*dx`

`I_1 = 1/2[tan x]_(-π/4)^(π/4)`

`I_1 = 1/2[1 - (-1)]`

∴ I1 = 1

Now,

`I_2 = int_(-1/2)^(1/2) |x|dx`

`I_2 = 2 int_0^(1/2) x*dx`

`I_2 = 2 [x^2/2]_0^(1/2)`

∴ `I_2 = 1/4`

So,

`I_1 - 4I_2 = 1 - 4(1/4)`

= 1 − 1

= 0

Hence proved.

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2025-2026 (March) 65/1/1
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