1
Question
If the value of the integral ∫cosx(1−sinx)(2−sinx)dx is ln∣∣∣a−sinxb−sinx∣∣∣+C, then a+b=
(where C is integration constant)
(where C is integration constant)
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Solution
Given: ∫cosx(1−sinx)(2−sinx)dx
Let sinx=t⇒cosx dx=dt
∴∫cosx(1−sinx)(2−sinx)dx=∫dt(1−t)(2−t)
Let
1(1−t)(2−t)=A(1−t)+B(2−t)⇒1=A(2−t)+B(1−t)⋯(1)
Equating the coefficients of t and constant term, we obtain
−A−B=02A+B=1
On solving, we obtain
A=1 and B=−1
∴∫1(1−t)(2−t) dt=∫[1(1−t)−1(2−t)] dt=−ln|1−t|+ln|2−t|+C=ln∣∣∣2−t1−t∣∣∣+C=ln∣∣∣2−sinx1−sinx∣∣∣+C∴a+b=3
Let sinx=t⇒cosx dx=dt
∴∫cosx(1−sinx)(2−sinx)dx=∫dt(1−t)(2−t)
Let
1(1−t)(2−t)=A(1−t)+B(2−t)⇒1=A(2−t)+B(1−t)⋯(1)
Equating the coefficients of t and constant term, we obtain
−A−B=02A+B=1
On solving, we obtain
A=1 and B=−1
∴∫1(1−t)(2−t) dt=∫[1(1−t)−1(2−t)] dt=−ln|1−t|+ln|2−t|+C=ln∣∣∣2−t1−t∣∣∣+C=ln∣∣∣2−sinx1−sinx∣∣∣+C∴a+b=3
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