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Question

0π/2a2 cos2 x+b2 sin2 x dx

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Solution

Let I=0π2a2 cos2 x+b2 sin2 x dx. Then, I=0π2a2 cos2 x+b2 1-cos2 x dxI=0π2b2+a2-b2 cos2 x dxI=0π2b2+a2-b21+cos 2x2dxI=b2x+a2-b22x+sin 2x20π2I=b2π2+a2-b22π2+0I=π4a2+b2

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