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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 ∫0π/4(sin⁡2x)dx\int\limits_{0}^{\pi/4} (\sin 2x) dx is:
Solution:  
Explanation: ∫0π/4sin⁡2xdx\int\limits_{0}^{\pi/4} \sin 2x dx
Let     u=2x\;\; u=2x
If x=0x=0 then, u=0u=0
   and x=π/4   then u=π/2  .   \;\text{ and } x=\pi/4 \;\text{ then } u=\pi/2 \;\text{. }\;
⇒    du=2dx\Rightarrow \;\; du=2 dx
  12∫0π/2sin⁡udu=−12[cos⁡u]0π/2\; \frac{1}{2} \int\limits_{0}^{\pi/2} \sin u du = -\frac{1}{2} [\cos u]_{0}^{\pi/2}
=−12[0−1]=12= -\frac{1}{2}[0-1] = \frac{1}{2}
$ $
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