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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.3
Find the integrals of the functions in Exercises 62 to 83.
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Question : 75 of 271
Marks: +1, -0
cos⁡x−sin⁡x1+sin⁡2x\frac{\cos x - \sin x}{1+\sin 2x}
Solution:  
Let I=∫cos⁡x−sin⁡x1+sin⁡2x dxI=\int \frac{\cos x - \sin x}{1+\sin 2x}\,dx
=∫cos⁡x−sin⁡x1+2sin⁡xcos⁡x dx=\int \frac{\cos x - \sin x}{1+2\sin x \cos x}\,dx
=∫cos⁡x−sin⁡xsin⁡2x+cos⁡2x+2sin⁡xcos⁡x dx=\int \frac{\cos x - \sin x}{\sin^2 x + \cos^2 x + 2\sin x \cos x}\,dx
=∫cos⁡x−sin⁡x(cos⁡x+sin⁡x)2 dx=\int \frac{\cos x - \sin x}{(\cos x + \sin x)^2}\,dx
Put cos⁡x+sin⁡x=t\cos x + \sin x = t
⇒(−sin⁡x+cos⁡x) dx=dt\Rightarrow (-\sin x + \cos x)\,dx = dt
∴I=∫dtt2=t−2+1−2+1+C\therefore I = \int \frac{dt}{t^2} = \frac{t^{-2+1}}{-2+1} + C
=−1t+C=-\frac{1}{t} + C =−1cos⁡x+sin⁡x+C=-\frac{1}{\cos x + \sin x} + C
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