Question

\(\int \frac{sinx}{sin4x}dx =\)

• A. \(\frac{1}{4\sqrt{2}}log\frac{1+\sqrt{2}sinx}{1-\sqrt{2}sinx}+\frac{1}{4}log|secx+tanx|+c\)
• B. \(\frac{1}{4\sqrt{2}}log\frac{1+\sqrt{2}sinx}{1-\sqrt{2}sinx}-\frac{1}{4}log|secx+tanx|+c\)
• C. \(\frac{1}{4\sqrt{2}}log\frac{1-\sqrt{2}sinx}{1+\sqrt{2}sinx}+\frac{1}{4}log|secx+tanx|+c\)
• D. \(\frac{1}{4\sqrt{2}}log\frac{1-\sqrt{2}sinx}{1+\sqrt{2}sinx}-\frac{1}{4}log|secx+tanx|+c\)

Hint:

Write sin4x = 4 sinx cosx cos2x, put t = sinx and split into partial fractions.

Answer 1

The Correct Option is B

Step 1: Simplify the integrand:
Use \(\sin 4x = 2\sin 2x\cos 2x = 4\sin x\cos x\cos 2x\). Then \[ \frac{\sin x}{\sin 4x} = \frac{1}{4\cos x\cos 2x} \]

Step 2: Prepare the substitution:
Multiply the top and bottom by \(\cos x\) and write \(\cos 2x = 1 - 2\sin^2 x\): \[ I = \frac{1}{4}\int \frac{\cos x\,dx}{\cos^2 x\,(1 - 2\sin^2 x)} = \frac14 \int \frac{\cos x\,dx}{(1-\sin^2 x)(1 - 2\sin^2 x)} \]

Step 3: Substitute:
Let \(t = \sin x\), so \(dt = \cos x\,dx\): \[ I = \frac14 \int \frac{dt}{(1-t^2)(1-2t^2)} \]

Step 4: Partial fractions:
Write \(\frac{1}{(1-t^2)(1-2t^2)} = \frac{A}{1-t^2} + \frac{B}{1-2t^2}\). Then \(A(1-2t^2) + B(1-t^2) = 1\), which gives \(A + B = 1\) and \(2A + B = 0\). So \(A = -1\) and \(B = 2\). \[ I = \frac14\left[-\int\frac{dt}{1-t^2} + 2\int\frac{dt}{1-2t^2}\right] \]

Step 5: Integrate:
Use \(\int\frac{dt}{1-t^2} = \frac12\log\frac{1+t}{1-t}\) and \(\int\frac{dt}{1-2t^2} = \frac{1}{2\sqrt2}\log\frac{1+\sqrt2 t}{1-\sqrt2 t}\). Then \[ I = \frac{1}{4\sqrt2}\log\frac{1+\sqrt2 t}{1-\sqrt2 t} - \frac18\log\frac{1+t}{1-t} + c \]

Step 6: Convert back:
With \(t = \sin x\), \(\frac12\log\frac{1+\sin x}{1-\sin x} = \log|\sec x + \tan x|\). So the second term is \(-\frac14\log|\sec x + \tan x|\).

Step 7: Match the options:
The result has \(+\frac{1}{4\sqrt2}\log\frac{1+\sqrt2\sin x}{1-\sqrt2\sin x}\) and \(-\frac14\log|\sec x+\tan x|\). Options (C) and (D) have the ratio inverted, which flips the sign of the first term. Option (A) has the wrong sign on the second term. Only (B) matches.

Final Answer:
The integral equals the expression in option (B). \[ \boxed{\frac{1}{4\sqrt2}\log\frac{1+\sqrt2\sin x}{1-\sqrt2\sin x}-\frac14\log|\sec x+\tan x|+c} \]