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Evaluate the following $\int\sqrt{1+\sin x}\;dx$

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  • $\sin ^2 \theta+\cos ^2 \theta=1$
  • $\sin 2 \theta=2 \sin \theta \cos \theta$
  • $\int \sin x dx=-\cos x+c$
Let $I=\int\sqrt{1+\sin x}\;dx$
$\sin ^2 x/2+\cos ^2 x/2=1$ and $\sin x=2 \sin x/2 \cos x/2$
substituting this in I we get
Hence $I=\int \sqrt {\sin^2 x/2+\cos ^2 x/2+2 \sin x/2 \cos x/2}\; dx$
But $\sin^2 x/2+\cos ^2 x/2+2 \sin x/2 \cos x/2=(\sin x/2+\cos x/2)^2$
Therefore $I=\int \sqrt {(\sin x/2+\cos x/2)^2 dx}$
$I=\int (\sin x/2+\cos x/2) dx$
$I=\int \sin.x/2 dx+\int \cos x/2 dx$
on integrating we get
$\large\frac{1}{\frac{1}{2}}(-\cos \frac{x}{2})+\frac{1}{\frac{1}{2}}\sin \frac{x}{2}+c$
$=2[\sin \large\frac{x}{2}-\cos \frac{x}{2}]+c$
answered Apr 11, 2013 by meena.p