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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 : 258 of 271
Marks: +1, -0
∫0π/4sin⁡x+cos⁡x9+16sin⁡2xdx\int\limits_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}dx
Solution:  
Let I=∫0π/4sin⁡x+cos⁡x9+16sin⁡2xdxI=\int\limits_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}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
and 1−2sin⁡xcos⁡x=t21 - 2 \sin x \cos x = t^2 ⇒1−sin⁡2x=t2\Rightarrow 1 - \sin^2 x = t^2
When x=π/4x = \pi/4, t=sin⁡π4−cos⁡π4t = \sin\frac{\pi}{4} - \cos\frac{\pi}{4} =12−12=0= \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0
t=sin⁡0−cos⁡0=−1t = \sin 0 - \cos 0 = -1
∴∫0π/4sin⁡x+cos⁡x9+16sin⁡2xdx\therefore \int\limits_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}dx
=∫−10dt9+16(1−t2)= \int\limits_{-1}^{0}\frac{dt}{9+16(1-t^2)}
=∫−10dt25−16t2= \int\limits_{-1}^{0}\frac{dt}{25-16t^2}
=116∫−10dt(54)2−t2= \frac{1}{16}\int\limits_{-1}^{0}\frac{dt}{\left(\frac{5}{4}\right)^2 - t^2}
=116⋅12×54[log⁡∣54+t54−t∣]−10= \frac{1}{16} \cdot \frac{1}{2 \times \frac{5}{4}} \left[ \log \left| \frac{\frac{5}{4}+t}{\frac{5}{4}-t} \right| \right]^0_{-1}
=140[log⁡1−(log⁡1−log⁡9)]= \frac{1}{40} [ \log 1 - (\log 1 - \log 9) ]
=140log⁡9= \frac{1}{40} \log 9
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