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∫0π4sin2xsin4x+cos4xdx = ____________ -

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Question

`int_0^{pi/4} (sin2x)/(sin^4x + cos^4x)dx` = ____________

Options

  • `pi/2`

  • `pi/4`

  • π

MCQ
Fill in the Blanks

Solution

`int_0^{pi/4} (sin2x)/(sin^4x + cos^4x)dx` = `underline(pi/4)`.

Explanation:

Let I = `int_0^{pi/4} (sin2x)/(sin^4x + cos^4x)dx`

= `int_0^{pi/4} (2sinx cosx)/(sin^4x + cos^4x)dx`

= `int_0^{pi/4} (2tanx sec^2x)/(tan^4x + 1)dx`

Put tan2x = t ⇒ 2tanx sec2x dx = dt

∴ I = `int_0^1 dt/(t^2 + 1)`

= `[tan^-1t]_0^1`

= `tan^-1 (1) - tan^-1(0)`

= `pi/4 - 0`

∴ I = `pi/4`

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