Question

integration question
\[\int\limits_{o}^{\pi/2}\frac{1}{1+sinx} dx =1\]

Answer 1

\[\int\limits_{o}^{\pi/2}\frac{1}{1+sinx} dx =1\]

Answer 2

I got \[[tanx-secx]_{0}^{\pi/2}\] But don't know how to evaluate it :|

Answer 3

How did you evaluate the integral?

Answer 4

Sub. the number but tan (pi/2) is undefined... so is sec(pi/2)

Answer 5

I mean how did you do the integration?
I differentiated the last but not getting 1/(1+sin x)

Answer 6

\[\int\frac{1}{1+sinx}dx = \int\frac{1-sinx}{1-sin^2x}dx\]\[ = \int\frac{1-sinx}{cos^2x}dx = \int sec^2x - tanxsecxdx\]\[ =tanx - secx +C\]

Answer 7

In this case... not C...Sorry.

Answer 8

But when I sub it, as I said, it is undefined.
And I haven't learnt improper integral yet

Answer 9

\[\int\limits_{0}^{\frac{\pi}{2}} \frac{1}{1 + \sin x}dx = \left . \frac{2 \sin \left( \frac{x}{2} \right)}{\sin \left( \frac{x}{2} \right) + \cos \left( \frac{x}{2} \right)}\right|_0^{\frac{\pi}{2}}\]

Answer 10

use t-method.
t=tan(x/2)
sinx=2t/(1-t^2)
dx=(2/(1-t^2))

int (1/(1+sinx))dx
=int (2/(1+2t+t^2))dt
=int (2/(t+1)^2)dt
=2 int(1/(t+1)^2)d(t+1)
=2int(t+1)^(-2) d(t+1)
=2 (-1)(t+1)^(-1)

when x=pi/2
t=tan(pi/4)=1
when x=0
t=tan0=0

therefore,
int (1/(1+sinx))dx
=2 (-1)(1+1)^(-1) - 2 (-1)(0+1)^(-1)
=-1+2
=1

Answer 11

We'll use limits here
We have
\[[\tan x-\sec x]_{0} ^{\pi/2}\]
Lower limit is just -1
\[\lim_{x \to \pi/2 } \frac{\sin x-1}{\cos x}-(-1)\]
write \[x= \pi/2+ h\], we'll change limits here \(h\to 0\), so
\[\lim_{h \to 0} \frac{ \cos h-1}{\sin h} -(-1)\]
Can you proceed now?

Answer 12

\[\tan x- \sec= \frac{\sin x-1}{\cos x}\]

Answer 13

sorry, denominator of sinx and dt should be 1+t^2

Answer 14

just these two is wrong. The other calculations are correct

Answer 15

\[\lim_{h \to 0} \frac{-2\sin (h^2/2)}{2 \sin h/2 \cos h/2}=\lim_{h \to 0}\frac{-\sin h/2}{\cos h/2}= \frac{0}{1}=0\]
so we get
\[0-(-1)=1\]

Answer 16

\[\cos h-1= -2\sin^2 h/2\]
\[\sin h= 2\cos h/2 \times \sin h/2\]

Answer 17

Did you get it?
I have messed up a little

Answer 18

Whenever during evaluation of limits, you are getting indefinite forms. Apply limits to evaluate that

Answer 19

I think I understand it... Thanks!

Answer 20

Welcome:D