∫cos⁡x−sin⁡x1+sin⁡2xdx is equal to - Infinity Learn by Sri Chaitanya

A
$\sin x + \cos x + C$
B
$- (\sin x + \cos x) + C$
C
$- \frac{1}{(\sin x + \cos x)} + C$
D
None of the above

Solution:

$\begin{array}{l} Let I = \int \frac{\cos x - \sin x}{1 + \sin 2 x} dx \\ = \int \frac{\cos x - \sin x}{\sin^{2} x + \cos^{2} x + 2 \sin xcos x} dx \\ [∵ \sin^{2} x + \cos^{2} x = 1; \sin 2 x = 2 \sin xcos x] \\ = \int \frac{\cos x - \sin x}{(\sin x + \cos x)^{2}} dx \end{array}$

Let $\cos x + \sin x = t \Rightarrow - \sin x + \cos x = \frac{dt}{dx}$

$\begin{array}{l} \Rightarrow dx = \frac{dt}{(\cos x - \sin x)} \\ ∴ I = \int \frac{\cos x - \sin x}{t^{2}} \cdot \frac{dt}{(\cos x - \sin x)} = \int \frac{1}{t^{2}} dt = \int t^{- 2} dt \\ = \frac{t^{- 2 + 1}}{- 2 + 1} + C = \frac{- 1}{\cos x + \sin x} + C \end{array}$

$\int \frac{\cos x - \sin x}{1 + \sin 2 x} dx$ is equal to

Solution:

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