Question
Evaluate: $\int_0^{\frac{\pi}{2}} \frac{\cos x}{(1+\sin x)(2+\sin x)} d x$
✓
Answer
Let $I =\int_0^{\frac{\pi}{2}} \frac{\cos x}{(1+\sin x)(2+\sin x)} d x$
Put $\sin x=t$
$\therefore \cos xdx = dt$
When $x =0, t =0$ and when $x =\frac{\pi}{2}, t =1$
$\therefore I=\int_0^1 \frac{ dt }{(1+ t )(2+ t )}$
Let $\frac{1}{(1+t)(2+t)}=\frac{A}{1+t}+\frac{B}{2+t} \ldots \ldots \ldots$ (i)
$\therefore 1= A (2+ t )+ B (1+ t )\ldots(ii)$
Putting $t=-1$ in (ii), we get
$A=1$
Putting $t=-2$ in (ii), we get
$1=- B$
$\therefore B =-1$
From (i), we get
$ \frac{1}{(1+ t )(2+ t )}=\frac{1}{1+ t }-\frac{1}{2+ t }$
$\therefore I =\int_0^1\left(\frac{1}{1+ t }-\frac{1}{2+ t }\right) dt$
$=\int_0^1 \frac{1}{1+ t } dt -\int_0^1 \frac{1}{2+1} dt$
$=[\log |1+ t |]_0^1-[\log |2+ t |]_0^1$
$=(\log 2-\log 1)-(\log 3-\log 2)$
$=\log 2-0-\log \left(\frac{3}{2}\right)$
$=\log \left(2 \times \frac{2}{3}\right)$
$\therefore I=\log \left(\frac{4}{3}\right) $
Put $\sin x=t$
$\therefore \cos xdx = dt$
When $x =0, t =0$ and when $x =\frac{\pi}{2}, t =1$
$\therefore I=\int_0^1 \frac{ dt }{(1+ t )(2+ t )}$
Let $\frac{1}{(1+t)(2+t)}=\frac{A}{1+t}+\frac{B}{2+t} \ldots \ldots \ldots$ (i)
$\therefore 1= A (2+ t )+ B (1+ t )\ldots(ii)$
Putting $t=-1$ in (ii), we get
$A=1$
Putting $t=-2$ in (ii), we get
$1=- B$
$\therefore B =-1$
From (i), we get
$ \frac{1}{(1+ t )(2+ t )}=\frac{1}{1+ t }-\frac{1}{2+ t }$
$\therefore I =\int_0^1\left(\frac{1}{1+ t }-\frac{1}{2+ t }\right) dt$
$=\int_0^1 \frac{1}{1+ t } dt -\int_0^1 \frac{1}{2+1} dt$
$=[\log |1+ t |]_0^1-[\log |2+ t |]_0^1$
$=(\log 2-\log 1)-(\log 3-\log 2)$
$=\log 2-0-\log \left(\frac{3}{2}\right)$
$=\log \left(2 \times \frac{2}{3}\right)$
$\therefore I=\log \left(\frac{4}{3}\right) $
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