Question

please solve this
Integration question .​

Answer 1

Answer:

ln ( | 1 + sin ( x ) )|/(| 2 + sin ( x ) | ) + C

or

log 1+sinx/2+sinx +c

OPTION (A) WILL BE CORRECT

Answer 2

Answer:

$\huge\orange{\overbrace{\red{\underline{\pink{\red\:{Answer}}}}}}$

= ㏒${\huge{|}}{\frac{sinx\;+\;1}{sinx\;+\;2}}{\huge{|}}+ c$

Step-by-step explanation:

Given that :-

I =  $\int\frac{cosx}{(1+sinx)(2+sinx)} dx$

Taking t = sinx

We get dt = cos x dx

It can be written as

I = $\int\frac{dt}{(1+t)(2+t)}$

Taking

$\frac{1}{(1+t)(2+t)} =\frac{A}{1+t} +\frac{B}{2+t}$

We know that

A(2 + t) + B(1 + t) = 1

Taking t + 1 = 0 we get t = -1

Here A(2 - 1) + B(0) = 1

Where A = 1

Taking t + 2 = 0 we get t = -2

Here A (0) + B(-2 + 1) = -2

$\sf\red{By\;substituting\;the\;values:-}$

$\frac{1}{(1+t)(2+t)} =\frac{1}{1+t} +\frac{-1}{2+t}$

We can write it as :-

$\int\frac{1}{(1+t)(2+t)}dt=\int\frac{1}{1+t} dt-\int\frac{1}{2+t}dt$

By integrating w.r.tt :-

= ㏒  ${\large{|}}{1+t}{\large{|}}-$  ㏒ ${\large{|}}{t+2}{\large{|}}+c$

So we get

= ㏒${\large{|}}{\frac{t\;+\;1}{t\;+\;2}}{\large{|}}+ c$

$\sf\red{By\;substituting\;the\;values\; of\;t:-}$

= $\int\frac{cosx}{(1+sinx)(2+sinx)}dx$

= ㏒${\large{|}}{\frac{sinx\;+\;1}{sinx\;+\;2}}{\large{|}}+ c$

$\large\rm\green{Therefore\;option\;A\;is\;correct}$

$\huge\sf\blue{\bigstar\;Hope\;its\;helpful\;to\;you\;dear\;\bigstar}$