Question

Evaluate \[ \int \frac{\cos x+\sin2x}{1-\sin x-2\sin^2x}\,dx \]

• A. \(\frac13\log\left(\frac{(1-2\sin x)^2}{|1+\sin x|}\right)+c\)
• B. \(\frac13\log\left(\frac{|1+\sin x|^{-1}}{(1-2\sin x)^2}\right)+c\)
• C. \(\frac13\log\left(\frac{(1+2\sin x)^2}{|1-\sin x|}\right)+c\)
• D. \(\frac13\log\left(\frac{|1-\sin x|}{(1+2\sin x)^2}\right)+c\)

Hint:

When both \(\sin x\) and \(\cos x\) appear together, substitution \(t=\sin x\) usually simplifies the integral immediately.

Answer 1

The Correct Option is A

Concept: Use substitution when denominator contains trigonometric polynomial. Take \[ t=\sin x \] Then \[ dt=\cos x\,dx \] Also \[ \sin2x=2\sin x\cos x \]

Step 1: Rewrite integral.
\[ I= \int \frac{\cos x+2\sin x\cos x} {1-\sin x-2\sin^2x} dx \] Factor numerator. \[ = \int \frac{\cos x(1+2\sin x)} {1-\sin x-2\sin^2x} dx \] Substitute \[ t=\sin x \] \[ I= \int \frac{1+2t}{1-t-2t^2}\,dt \]

Step 2: Factor denominator.
\[ 1-t-2t^2 = (1-2t)(1+t) \] Thus \[ I= \int \frac{1+2t}{(1-2t)(1+t)}dt \] Partial fractions give \[ = \frac23\frac1{1-2t} + \frac13\frac1{1+t} \]

Step 3: Integrate.
\[ I = -\frac13\log|1-2t| + \frac13\log|1+t| \] Substituting back \[ I= \frac13 \log \left( \frac{(1-2\sin x)^2}{|1+\sin x|} \right)+c \] Hence \[ \boxed{ \frac13\log \left( \frac{(1-2\sin x)^2}{|1+\sin x|} \right)+c } \]