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Evaluate the following $\int\frac{(\sin x+\cos x)}{\sqrt {1+\sin 2x}}dx$

$\begin{array}{1 1}(A)\;x+c \\ (B)\;\frac{x^2}{2}+c \\ (C)\;-x+c \\ (D)\;-\frac{x^2}{2}+c \end{array} $

1 Answer

  • $\cos ^2 x+\sin ^2x=1$
  • $\int dx=x+c$
Let $I=\large\int\frac{(\sin x+\cos x)}{\sqrt {1+\sin 2x}}dx$
we know $\cos ^2x+\cos x=1$ and $\sin 2x=2 \sin x \cos x$
Substituting this in I we get,
Therefore $I=\large\int\frac{(\sin x+\cos x)}{\sqrt {\sin^ 2x+\cos ^2 x+2 \sin x \cos x}}dx$
$\sin ^ 2x+\cos ^2 x+2 \sin x \cos x=(\sin x+\cos x)^2$
Hence $I=\large\int\frac{(\sin x+\cos x)}{\sqrt {(\sin x+\cos x)^2}}dx$
$I=\large\int\frac{\sin x+\cos x}{\sin x+\cos x}dx=\int dx$
On integrating we get,
answered Apr 11, 2013 by meena.p