logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

Evaluate the definite integrals\[\int\limits_\frac{\pi}{6}^\frac{\pi}{3}\frac{\sin x+\cos x}{\sqrt{\sin2x}}dx\]

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • (i)when f(x) is an integral function is substitute on t,then $f'(x)dx=dt.Hence\;\int f(x)dx=\int t.dt$
  • (ii)$\sin 2x=2\sin x \cos x$
  • (ii)$\int \frac{dx}{a^2-x^2}=\sin ^{-1}(x/a)+c$
  • (iv) if f(x) is an even function, then $\int \limits_{-1}^a f(x)dx=2\int \limits _0^a f(x)dx$
Given $I=\int\limits_\frac{\pi}{6}^\frac{\pi}{3}\frac{\sin x+\cos x}{\sqrt{\sin2x}}dx$ But $\sin 2x=2\sin x\cos x$
$=\int\limits_\frac{\pi}{6}^\frac{\pi}{3}\frac{\sin x+\cos x}{\sqrt{2\sin x \cos x}}dx$
Add and subtract 1 to the determinator,
$\int\limits_\frac{\pi}{6}^\frac{\pi}{3}\frac{\sin x+\cos x}{\sqrt{1-1+2\sin x \cos x}}dx$
we can write this as
$\int\limits_\frac{\pi}{6}^\frac{\pi}{3}\frac{\sin x+\cos x}{1-(1-2\sin x \cos x)}dx$
But $1=\sin ^2 x+\cos ^2x$
$\large\int\limits_\frac{\pi}{6}^\frac{\pi}{3}\frac{\sin x+\cos x}{1-(\sin^2x+\cos ^2x-2\sin x \cos x)}dx$
$\sin ^2x+\cos ^2x-2\sin x \cos x=(\sin x-\cos x)^2$
Therefore $I=\int\limits_\frac{\pi}{6}^\frac{\pi}{3}\frac{\sin x+\cos x}{\sqrt {1-(\sin x \cos x)^2}}dx$
Let $\sin x-\cos x=t;$ on differentiating w.r.t x we get,
$(\cos x+\sin x)dx=dt$
But the limit also change when we substitute for t
When $x=\frac{\pi}{6};\qquad \sin x-\cos x=t$
$=>\sin \pi/6-\cos \pi/6 =\frac{1-\sqrt 3}{2}=t$
When $x=\pi/3; \sin \pi/3 -\cos \pi/3=\frac{\sqrt 3 -1}{2}=t$
Therefore $I=\int \limits_{\frac{-\sqrt 3-1}{2}}^{\frac{\sqrt 3-1}{2}} \frac{dt}{\sqrt {1-t^2}}$
This is the form of $\int \frac{dx}{\sqrt {a^2-x^2}}=2\sin ^{-1}(x/a)+c$
Hence $I=\int \limits_{\frac{-\sqrt 3-1}{2}}^{\frac{\sqrt 3-1}{2}} \frac{dt}{\sqrt {1-t^2}}$
If we replcae 't' by -t, we get back.
$\sqrt {1-t^2},$ hence it is an even function
Therefore $I=2\int \limits_{\frac{-\sqrt 3-1}{2}}^{\frac{\sqrt 3-1}{2}} \frac{dt}{\sqrt {1-t^2}}$
On integrating we get $2[\sin^{-1}t]_0^{\frac{\sqrt 3-1}{2}}+c$
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}(\frac{\sqrt 3-1}{2})+c$
answered Feb 19, 2013 by meena.p
edited Feb 5, 2014 by rvidyagovindarajan_1
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...