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Question

0π/2sin x-cos x dx

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Solution

0π2sinx-cosxdx=20π2sinx12-cosx12dx=20π2sinx cosπ4-cosx sinπ4dx=20π2sinx-π4dxWe have,sinx-π4=-sinx-π4, 0xπ4 sinx-π4, π4xπ20π2sinx-cosxdx=20π4-sinx-π4dx+2π4π2sinx-π4dx =2cosx-π40π4-2cosx-π4π4π2 =2cos 0-cos-π4-2cosπ4-cos 0 =21-12-12+1 =22-22 = 22-2 = 22-1

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