Question

If \[ \int \frac{dx}{\sin^3x+\cos^3x} = A\log\left| \frac{\sqrt2+t}{\sqrt2-t} \right| + B\tan^{-1}(t)+C, \] then \[ \left(\frac{B}{A},\,t\right) = \]

• A. \[ \left(2\sqrt2,\;\sin x+\cos x\right) \]
• B. \[ \left(\frac{\sqrt2}{9},\;\sin x+\cos x\right) \]
• C. \[ \left(\frac{\sqrt2}{9},\;\sin x-\cos x\right) \]
• D. \[ \left(2\sqrt2,\;\sin x-\cos x\right) \]

Hint:

For integrals containing \[ \sin^3x+\cos^3x, \] the substitution \[ t=\sin x-\cos x \] is highly effective because \[ (\sin x+\cos x)^2=2-t^2 \] and \[ \sin x\cos x=\frac{1-t^2}{2}. \]

Answer 1

The Correct Option is D

Step 1: Factor the denominator. \[ \sin^3x+\cos^3x = (\sin x+\cos x) (\sin^2x-\sin x\cos x+\cos^2x). \] Since \[ \sin^2x+\cos^2x=1, \] \[ \sin^3x+\cos^3x = (\sin x+\cos x)(1-\sin x\cos x). \] Let \[ t=\sin x-\cos x. \] Then \[ t^2 = 1-2\sin x\cos x. \] Hence \[ \sin x\cos x=\frac{1-t^2}{2}. \] Also, \[ (\sin x+\cos x)^2 = 1+2\sin x\cos x = 2-t^2. \] Thus \[ \sin x+\cos x=\sqrt{2-t^2}. \] and \[ 1-\sin x\cos x = \frac{1+t^2}{2}. \] Therefore \[ \sin^3x+\cos^3x = \frac{\sqrt{2-t^2}\,(1+t^2)}{2}. \]

Step 2: Find \(dt\). \[ t=\sin x-\cos x. \] \[ dt=(\cos x+\sin x)\,dx. \] \[ dt=\sqrt{2-t^2}\,dx. \] \[ dx=\frac{dt}{\sqrt{2-t^2}}. \] Substituting, \[ I = \int \frac{dx}{\sin^3x+\cos^3x} = \int \frac{2\,dt} {(1+t^2)(2-t^2)}. \]

Step 3: Use partial fractions. \[ \frac{2}{(1+t^2)(2-t^2)} = \frac{2/3}{1+t^2} + \frac{2/3}{2-t^2}. \] Hence \[ I = \frac23\int\frac{dt}{1+t^2} + \frac23\int\frac{dt}{2-t^2}. \] \[ = \frac23\tan^{-1}t + \frac23\cdot\frac1{2\sqrt2} \log\left| \frac{\sqrt2+t}{\sqrt2-t} \right| +C. \] Therefore \[ A=\frac1{3\sqrt2}, \qquad B=\frac23. \]

Step 4: Compute \(\dfrac{B}{A}\). \[ \frac{B}{A} = \frac{\frac23} {\frac1{3\sqrt2}} = 2\sqrt2. \] Also, \[ t=\sin x-\cos x. \]

Step 5: Write the final answer. \[ \boxed{\left(2\sqrt2,\;\sin x-\cos x\right)} \]