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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 : 255 of 271
Marks: +1, -0
∫0π/2cos⁡2xcos⁡2x+4sin⁡2xdx\int\limits_{0}^{\pi/2} \frac{\cos^2 x}{\cos^2 x+4\sin^2 x}dx
Solution:  
Let I=∫0π/2cos⁡2xcos⁡2x+4sin⁡2xdxI=\int\limits_{0}^{\pi/2} \frac{\cos^2 x}{\cos^2 x+4\sin^2 x}dx
=∫0π/2cos⁡2xcos⁡2x+4(1−cos⁡2x)dx=\int\limits_{0}^{\pi/2} \frac{\cos^2 x}{\cos^2 x+4(1-\cos^2 x)}dx
=∫0π/2cos⁡2x4−3cos⁡2xdx=\int\limits_{0}^{\pi/2} \frac{\cos^2 x}{4-3\cos^2 x}dx
=−13∫0π/24−cos⁡2x−44−3cos⁡2xdx=-\frac{1}{3}\int\limits_{0}^{\pi/2} \frac{4-\cos^2 x-4}{4-3\cos^2 x}dx
=−13(π2)+43∫0π/2sec⁡2x4sec⁡2x−3dx=-\frac{1}{3}\left(\frac{\pi}{2}\right)+\frac{4}{3}\int\limits_{0}^{\pi/2} \frac{\sec^2 x}{4\sec^2 x-3}dx
=−π6+43∫0π/2sec⁡2x4(1+tan⁡2x)−3dx=-\frac{\pi}{6}+\frac{4}{3}\int\limits_{0}^{\pi/2} \frac{\sec^2 x}{4(1+\tan^2 x)-3}dx
Put tan⁡x=t\tan x = t ⇒sec⁡2xdx=dt\Rightarrow \sec^2 x dx = dt
When x=0,t=0x = 0, t = 0 and when x=π/2,t=∞x = \pi/2, t = \infty
∴I=−π6+43∫0∞dt4(1+t2)−3\therefore I = -\frac{\pi}{6} + \frac{4}{3} \int\limits_{0}^{\infty} \frac{dt}{4(1+t^2)-3}
=−π6+43∫0∞dt4t2+1= -\frac{\pi}{6} + \frac{4}{3} \int\limits_{0}^{\infty} \frac{dt}{4t^2+1}
=−π6+43⋅14∫0∞dtt2+14= -\frac{\pi}{6} + \frac{4}{3} \cdot \frac{1}{4} \int\limits_{0}^{\infty} \frac{dt}{t^2+\frac{1}{4}}
=−π6+13⋅112[tan⁡−1t12]0∞= -\frac{\pi}{6} + \frac{1}{3} \cdot \frac{1}{\frac{1}{2}} \left[ \tan^{-1} \frac{t}{\frac{1}{2}} \right]_{0}^{\infty}
=−π6+23[tan⁡−12t]0∞= -\frac{\pi}{6} + \frac{2}{3} \left[ \tan^{-1} 2t \right]_{0}^{\infty}
=−π6+23[tan⁡−1∞−tan⁡−10]= -\frac{\pi}{6} + \frac{2}{3} \left[ \tan^{-1} \infty - \tan^{-1} 0 \right]
=−π6+23[π2−0]= -\frac{\pi}{6} + \frac{2}{3} \left[ \frac{\pi}{2} - 0 \right]
=−π6+π3=π6= -\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6}
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