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Question

π013+2 sin x+cos xdx=

A
π3
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B
π4
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C
π6
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D
π2
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Solution

The correct option is B π4
Put t=tan x2. Then dt=sec2 x2.12dxdx=2dt1+t2; x=0, πt=0, sin x=2t1+t2, cos x=1t21+t2
π013+2 sin x+cos x dx=013+2⎢ ⎢ ⎢ ⎢2t(1+t2)+[(1t2)(1+t2)]⎥ ⎥ ⎥ ⎥2dt1+t2
=0 1+t23+3t2+4t+1t22dt1+t2=2012t2+4t+4dt=01t2+2t+2dt
=01(t+1)2+12dt=[Tan1(t+1)]0=π2π4=π4

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