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∫0π4(xxsinx+cosx)2dx=

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a

5−π5+π

b

24+π

c

4−π4+π

d

4+π4−π

answer is C.

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Detailed Solution

∫0π4x2dx(xsinx+cosx)2=∫0π4xsecx.xcosxdx(xsinx+cosx)2 =∫0π4xsecxddx(−1(xsinx+cosx))dx =−xsecxxsinx+cosx|0π4+∫0π4secx+xsecxtanxxsinx+cosxdx =−2ππ+4+∫0π4sec2xdx=−2ππ+4+1 =4−π4+π

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