Integration

Question

$\int_{0}^{\frac{\pi }{2}}\frac{cosx}{2\sqrt{1-sinx}}dx=?$

Comments

so what is mistake  here you got it ? iam unable to find

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I appreciated ur approach :)

bariya means very good :p

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ha janta hu mtlb  bariya ka :P  iam asking correct answer is 1 but  i  got -1  if i take   

1- sin x = (  sin (x/2 ) - cos(x/2) )^2

 and if i take 1-sinx =  ( cos(x/2) - sin(x/2) ) ^2   then answer =  +1

why ??  so   -1 should not come

Answer 1

$\int_{0}^{\frac{\pi }{2}} \frac{cos x }{2\sqrt{1 - sin x}} dx$

put sin x =t   , cos x dx = dt  , now   t will vary from 0 to 1

$\int_{0}^{1} \frac{dt}{2\sqrt{1-t}}$

$\int_{0}^{1} \frac{1}{2} (1-t)^{\frac{-1}{2}}$

now ,  $\frac{1}{2} (1-t)^{\frac{1}{2}} *(-)2$

-  $(1-t)^{\frac{1}{2}}$   range from 0 to 1 you will   get  -(-1 ) = 1

Comments

you missed the -ve sign in 2nd last line

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it's 1 correct...