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Question

lf f(x)=(x2+sin2x)1+x2sec2xdx and f(0)=0 then f(1) is equal to

A
1π4
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B
π41
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C
tan1π4
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D
π4tan1
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Solution

The correct option is A tan1π4
Given, f(x)=(x2+sin2x)1+x2sec2xdx
(x2sec2x+tan2x)1+x2dx
(x2(1+tan2x)+tan2x)1+x2dx

(x2+1)tan2x+x21+x2dx
tan2x+1+x21+x211+x2dx
tanxtan1x+c ,where c is a constant.
Since, f(0)=0 we put x=0 in the above equation to get c=0 .
Hence , f(1)=tan1tan11=tan1π4 is the answer.

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