Indefinite IntegrationHard
Question
The value of the integral 
dx is
dx is Options
A.sin x - 6 tan-1(sin x) + c
B.sin x - 2(sin x)-1 + c
C.sin x - 2(sin x)-1 - 6 tan-1(sin x) + c
D.sin x - 2(sin x)-1 + 5 tan-1(sin x) + c
Solution
Let I = dx
Put sin x = t
⇒ cos x dx = dt
∴ I =
dx
⇒
.....(i)
Using partial fraction for,
(where y = t2)
⇒ A = 2, B = -6
∴
∴ Eq. (i) reduces to,
-6 tan-1 (t) + c
= sin x -
- 6 tan-1 (sin x) + c
dxPut sin x = t
⇒ cos x dx = dt
∴ I =
dx ⇒
.....(i) Using partial fraction for,
(where y = t2) ⇒ A = 2, B = -6
∴
∴ Eq. (i) reduces to,
-6 tan-1 (t) + c = sin x -
- 6 tan-1 (sin x) + cCreate a free account to view solution
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