Question

Integrate cosx/[ (1-sinx) (2-sinx) ] with respect to x

Answer 1

Here, we need to evaluate ∫cosx(1−sinx)(2+sinx)dx

First put sinx = t ...(1)
Differentiating (1) with respect to x, we get
cosxdx=dt ...(2)

Now, substituting (1) and (2) in the given integral, we get
∫cosx(1−sinx)(2+sinx)dx=∫dt(1−t)(2+t)

Next, solve the above integral using the method of partial fractions.So,
1(1−t)(2+t)=A1−t+B2+t1(1−t)(2+t) =A(2+t)+B(1−t)(1−t)(2+t)1=A(2+t)+B(1−t)
1=2A+At+B−Bt
Now,
2A+B=1
A-B=0

Solving for A and B, we get
A=1/3
B=1/3

So,
A1−t+B2+t=(13)1−t+(13)2+t=(13)(11−t+12+t)
Thus,

∫cosx(1−sinx)(2+sinx)dx=∫(13)(11−t+12+t)dt=(13)∫(11−t+12+t)dt=(13)(log|1+t|+log|2+t|)=(13)(log|(1+t)(2+t)|)

Substituting t=sinx, we get

∫cosx(1−sinx)(2+sinx)dx=(13)(log|(1+sinx)(2+sinx)|)+C

Therefore,
∫cosx(1−sinx)(2+sinx)dx=(13)(log|(1+sinx)(2+sinx)|)+C