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NCERT Class XII Mathematics Chapter - - Solutions

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Evaluate the definite integrals in Exercises 253 to 261.
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Question : 256 of 271
Marks: +1, -0
∫π/6π/3sin⁡x+cos⁡xsin⁡2x dx\int\limits_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}\,dx
Solution:  
Let ∫=∫π/6π/3sin⁡x+cos⁡xsin⁡2x dx\int = \int\limits_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}\,dx
=∫π/6π/3sin⁡x+cos⁡x1−(1−sin⁡2x) dx= \int\limits_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{1-(1-\sin 2x)}}\,dx
=∫π/6π/3sin⁡x+cos⁡x1−(sin⁡x−cos⁡x)2 dx= \int\limits_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}}\,dx
Put sin⁡x−cos⁡x=t\sin x - \cos x = t ⇒(cos⁡x+sin⁡x) dx=dt\Rightarrow (\cos x + \sin x)\,dx = dt
When x=π/6x = \pi/6, t=sin⁡π6−cos⁡π6t = \sin\frac{\pi}{6} - \cos\frac{\pi}{6} =12−32= \frac{1}{2} - \frac{\sqrt{3}}{2}
When x=π/3,t=sin⁡π3−cos⁡π3x = \pi/3, t = \sin\frac{\pi}{3} - \cos\frac{\pi}{3} =32−12= \frac{\sqrt{3}}{2} - \frac{1}{2}
∴I=∫12−3232−12dt1−t2=[sin⁡−1t]12−3232−12\therefore I = \int\limits_{\frac{1}{2}-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}-\frac{1}{2}} \frac{dt}{\sqrt{1-t^2}} = \left[\sin^{-1}t\right]_{\frac{1}{2}-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}-\frac{1}{2}}
=sin⁡−1(32−12)−sin⁡−1(12−32)= \sin^{-1}\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right) - \sin^{-1}\left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)
=sin⁡−1(32−12)+sin⁡−1(32−12)= \sin^{-1}\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right) + \sin^{-1}\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)
=2sin⁡−1(12(3−1))= 2\sin^{-1}\left(\frac{1}{2}(\sqrt{3}-1)\right)
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