1all the delhi people and all india are requested to solve this question. any more question can be added but make sure to write teh complete question.
The question that said
[sin x + cos x] / √sin2x
this function had to be integrated with limits (pi/6) to (pi/3)
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7 Answers
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1the given is equalt to \int\frac{\sqrt{tanx}+\sqrt{cotx}}{\sqrt{2}}dx
this can be eva;uated easily by taking tanx=t and cotx=v
1Integral is \frac{1}{\sqrt{2}}\int_{\pi /6}^{\pi /3}{\frac{sinx+cosx}{\sqrt{sinxcosx}}dx
now put sinx - cosx = t \Rightarrow (cosx+sinx)dx=dt
\Rightarrow (1-sin2x) = t^{2} \Rightarrow sin2x = t^{2}-1
now substituting in integral we get \int_{\pi /6}^{\pi /3}{\frac{dt}{\sqrt{1-t^{2}}}} = sin -1(sinx - cosx) + c now substitute the limits u will get the ans.
39Kaustab the way you're doing it, it'll be quite long even though it's an NCERT example. The easier way is to apply the substitution Manmay has.
And Manmay, after changing the integral variable to t we change the limits too. Your answer won't be in terms of x, it will be in terms of t. Little silly mistake..lol.
1i thin manmay is right der limits may be kept same if we express ans at last in terms of x and not t
39Jo mujhe sikhaya gaya hai I am going as per that SR....well I don't know, lets see [12]
1@ pritish i did that for it would be easy, keeping limits same when we get final answer in terms of t we substitute t ' s value and use limits
changing limits would be ime taking.
but i should have done that here.
sorry for the mistake.