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Evaluate : $ \int_{\Large\frac{\pi}{6}}^{\Large\frac{\pi}{3}}\large\frac{\sin\: x + \cos \: x}{\sqrt{\sin\: 2x}}$$ dx$

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  • when f(x) is an integral function is substitute on t,then $f'(x)dx=dt.Hence\;\int f(x)dx=\int t.dt$
  • $\sin 2x=2\sin x \cos x$
  • $\int \large\frac{dx}{a^2-x^2}$$=\sin ^{-1}(x/a)+c$
  • if f(x) is an even function, then $\int \limits_{-1}^a f(x)dx=2\int \limits _0^a f(x)dx$
Step 1:
$I=\int\limits_{\Large\frac{\pi}{6}}^{\Large\frac{\pi}{3}}\large\frac{\sin x+\cos x}{\sqrt{\sin2x}}$$dx$
But $\sin 2x=2\sin x\cos x$
$=\int\limits_{\Large\frac{\pi}{6}}^{\Large\frac{\pi}{3}}\large\frac{\sin x+\cos x}{\sqrt{2\sin x \cos x}}$$dx$
Add and subtract 1 to the determinator,
$\int\limits_{\Large\frac{\pi}{6}}^{\Large\frac{\pi}{3}}\large\frac{\sin x+\cos x}{\sqrt{1-1+2\sin x \cos x}}$$dx$
Step 2:
we can write this as
$\int\limits_{\Large\frac{\pi}{6}}^{\Large\frac{\pi}{3}}\large\frac{\sin x+\cos x}{1-(1-2\sin x \cos x)}$$dx$
But $1=\sin ^2 x+\cos ^2x$
$\large\int\limits_{\Large\frac{\pi}{6}}^{\Large\frac{\pi}{3}}\large\frac{\sin x+\cos x}{1-(\sin^2x+\cos ^2x-2\sin x \cos x)}$$dx$
$\sin ^2x+\cos ^2x-2\sin x \cos x=(\sin x-\cos x)^2$
Therefore $I=\int\limits_{\Large\frac{\pi}{6}}^{\Large\frac{\pi}{3}}\large\frac{\sin x+\cos x}{\sqrt {1-(\sin x \cos x)^2}}$$dx$
Step 3:
Let $\sin x-\cos x=t;$
on integrating w.r.t x we get,
$(\cos x+\sin x)dx=dt$
Now substituting t and dt we get
$I=\int\limits_{\Large\frac{\pi}{6}}^{\Large\frac{\pi}{3}}\large\frac{dt}{\sqrt {1-t^2}}$
But the limit also change when we substitute for t
When $x=\large\frac{\pi}{6}$$;\qquad \sin x-\cos x=t$
$=>\sin \pi/6-\cos \pi/6 =\frac{1-\sqrt 3}{2}=t$
Therefore $I=\int \limits_{\large\frac{-\sqrt 3-1}{2}}^{\large\frac{\sqrt 3-1}{2}} \large\frac{dt}{\sqrt {1-t^2}}$
Step 4:
This is the form of $\int \large\frac{dx}{\sqrt {a^2-x^2}}$$=2\sin ^{-1}(x/a)+c$
Hence $I=\int \limits_{\Large\frac{-\sqrt 3-1}{2}}^{\Large\frac{\sqrt 3-1}{2}} \large\frac{dt}{\sqrt {1-t^2}}$
If we replace 't' by -t, we get back.
$\sqrt {1-t^2},$
hence it is an even function
Therefore $I=2\int \limits_{\Large\frac{-\sqrt 3-1}{2}}^{\Large\frac{\sqrt 3-1}{2}} \large\frac{dt}{\sqrt {1-t^2}}$
On integrating we get
$2[\sin^{-1}t]_0^{\Large\frac{\sqrt 3-1}{2}}+c$
On applying limits,
$I=2 \bigg[\sin^{-1} (\frac{\sqrt 3-1}{2})-\sin ^{-1}(0)\bigg]+c$
But $\sin^{-1}(0)=0$
Therefore $ I=2 \sin^{-1}(\large\frac{\sqrt 3-1}{2})$$+c$
answered Sep 30, 2013 by sreemathi.v
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