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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.11
By using the properties of definite integrals, evaluate the integrals in Exercises 208 to 225.
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Question : 217 of 271
Marks: +1, -0
∫−π2π2sin⁡2xdx\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x dx
Solution:  
Let I=∫−π2π2sin⁡2xdxI=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x dx
I=2∫0π2sin⁡2xdxI=2\int\limits_{0}^{\frac{\pi}{2}} \sin^2 x dx [ ∵sin⁡2x\because \sin^2 x is an even function]
=2∫0π2(1−cos⁡2x2)dx=2\int\limits_{0}^{\frac{\pi}{2}} \left( \frac{1-\cos 2x}{2} \right) dx
=[x−sin⁡2x0]0π2=\left[ x - \frac{\sin 2x}{0} \right]_{0}^{\frac{\pi}{2}}
⇒I=π2\Rightarrow I = \frac{\pi}{2}
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