Question

If \(\rm \int \sqrt{1 - sin 2x} \space dx\) = A sinx + B cosx + C, where 0 < x < \(\frac{\pi}{4}\), then which one of the following is correct?
1. A + B = 0
2. A + B - 2 = 0
3. A + B + 2 = 0
4. A + B - 1 = 0

Answer 1

Correct Answer - Option 2 : A + B - 2 = 0

Concept:

sin²x + cos²x = 1

\(\rm \int sin x dx = -cosx \)

\(\rm \int cos x dx = sin x \)

Calculation:

We have \(\rm \int \sqrt{1 - sin 2x} \space dx\) = A sinx + B cosx + C

⇒ \(\rm \int (\sqrt{sin^{2}x + cos^{2}x - 2sinx cosx} )dx\) = A sinx + B cosx + C

⇒ \(\rm \int (\sqrt{(sinx - cosx)^{2}})dx\) = A sinx + B cosx + C

⇒ \(\rm \int (|(sinx - cosx)|)dx\) = A sinx + B cosx + C     ----(i)

If 0 < x < \(\frac{\pi}{4}\), then sinx < cosx

⇒ |sinx - cosx| = -sinx + cosx    -----(ii)

Now from (i) and (ii), we get 

⇒ \(\rm \int (-\space sinx + cosx) \space dx\) = A sinx + B cosx + C

⇒ cosx + sinx + C = A sinx + B cosx + C

On comparing A = 1, B = 1 and C = 0

Hence, A + B - 2 = 0 is correct.