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Integrals

Question
CBSEENMA12036290

The value of -π2π2sin2x1+2xdx is:

  • π/4

  • π/8

  • π/2

Solution

A.

π/4

I = -π2π2sin2 x dx1 + 2x ... (i)Also, I = -π2π22x sin2 x dx1 + 2x .... (ii)Adding (i) and (ii)2I = -π2π2 sin2 xdx2I = 20π2sin2 x dx I= 0π2sin2 xdx .... (iii)I =  0π2cos2 xdx .... (iv)Adding (iii) and (iv)2I =    0π2dx = π2I = π4

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