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CBSE Class 12 Math 2023 Delhi Set 2 Solved Paper

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Question : 4 of 14
Marks: +1, -0
The value of 0π4(sin2x)dx is:
Solution:  
Explanation: 0π4sin2xdx
Let u=2x
If x=0 then, u=0
and x=π4 then u=π2.
du=2dx
120π2sinudu=12[cosu]0π2
=12[01]=12
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