Question

If \( \int \frac{\sin 2x}{(1+\sin x)(2+\sin x)} \, dx = a \log |1+\sin x| - b \log |2+\sin x| + c \), then the value of \( a \) and \( b \) is __________.

• A. \( a = -2, b = 4 \)
• B. \( a = 2, b = 4 \)
• C. \( a = -2, b = -4 \)
• D. \( a = 2, b = -4 \)

Hint:

For integrals involving \( \sin x \) and \( \cos x dx \), use \( t=\sin x \) and then apply partial fractions.

Answer 1

The Correct Option is C

Step 1: Rewrite the integrand.
We know that:
\[ \sin 2x = 2\sin x \cos x \]
So the given integral becomes:
\[ \int \frac{2\sin x \cos x}{(1+\sin x)(2+\sin x)}dx \]

Step 2: Use substitution.
Let:
\[ t = \sin x \]
Then:
\[ dt = \cos x \, dx \]

Step 3: Convert the integral in terms of \( t \).
\[ \int \frac{2t}{(1+t)(2+t)}dt \]

Step 4: Apply partial fractions.
Let:
\[ \frac{2t}{(1+t)(2+t)} = \frac{A}{1+t} + \frac{B}{2+t} \]
Multiplying by \( (1+t)(2+t) \), we get:
\[ 2t = A(2+t) + B(1+t) \]

Step 5: Compare coefficients.
\[ 2t = (2A+B) + (A+B)t \]
Comparing constant and coefficient of \( t \):
\[ 2A+B = 0 \]
\[ A+B = 2 \]
Solving these equations:
\[ A = -2,\quad B = 4 \]

Step 6: Integrate.
\[ \int \frac{2t}{(1+t)(2+t)}dt = \int \left(\frac{-2}{1+t}+\frac{4}{2+t}\right)dt \]
\[ = -2\log|1+t|+4\log|2+t|+c \]
Now substitute \( t = \sin x \):
\[ = -2\log|1+\sin x|+4\log|2+\sin x|+c \]

Step 7: Compare with given expression.
Given form is:
\[ a\log|1+\sin x|-b\log|2+\sin x|+c \]
So:
\[ a = -2,\quad -b = 4 \]
\[ b = -4 \]
Therefore:
\[ \boxed{a=-2,\; b=-4} \]