Question

prove
\int\limits^{\pi/2}_{0} \sqrt{1-\sin x)/(1+sinx)dx}=1

Answer 1

yep

Answer 2

i think dx is outside

Answer 3

\[\int\limits\limits^{\pi/2}_{0} \sqrt{1-\sin x)/(1+sinx)}dx=1\]

Answer 4

\(LHS=\int\limits^{\pi/2}_{0} \sqrt{(1-\sin x)/(1+sinx)}dx\)

\(=\large\int\limits^{\pi/2}_{0} \sqrt{\frac{1 - \sin^2 x}{(1+\sin x)^2}}dx\)

\( =\large\int\limits^{\pi/2}_{0} \frac{\cos x}{(1+\sin x)}dx\)

Let \((1+\sin x) = t\), then, \(\cos x dx = dt\)

So, \( \large\int\limits^{\pi/2}_{0} \frac{dt}{(t)} = \large [\ln t ]^{\pi/2}_{0}= [ln(1+\sin x)]^{\pi/2}_{0}\)

= \(\large ln2 - ln1 = ln2\)

Answer 5

hw is that =1 ?

Answer 6

@apoorvk , just need to change your limits of integration....

Answer 7

@dpaInc but then I switched back to 'x' terms, so why would we convert the limits?

Answer 8

hmm... ur right.

Answer 9

So so so... where are we wrong? Or rather, who's wrong - us or the book? -.-

Answer 10

the book is wrong... wolfram says it's apoors answer.

Answer 11

Aye! :D