Question

Evaluate \[ \int_{0}^{\pi/4} \frac{\cos x-\sin x}{9+5\sin 2x}\,dx. \]

• A. \[ \frac{1}{\sqrt5} \left( \tan^{-1}\sqrt{10} -\tan^{-1}\sqrt5 \right) \]
• B. \[ \frac{1}{2\sqrt5} \left( \tan^{-1}\frac{\sqrt{10}}{2} -\tan^{-1}\frac{\sqrt5}{2} \right) \]
• C. \[ \frac{1}{2\sqrt5} \left( \tan^{-1}\sqrt{10} +\tan^{-1}\sqrt5 \right) \]
• D. \[ \frac{1}{\sqrt5} \left( \tan^{-1}\sqrt{\frac52} +\tan^{-1}\frac{\sqrt5}{2} \right) \]

Hint:

Whenever the numerator contains \(\cos x-\sin x\), try the substitution \[ t=\sin x+\cos x, \] because \[ dt=(\cos x-\sin x)\,dx. \] Also remember \[ (\sin x+\cos x)^2=1+\sin 2x. \]

Answer 1

The Correct Option is B

Concept: Use the substitution \[ t=\sin x+\cos x. \] Since \[ dt=(\cos x-\sin x)\,dx, \] the numerator becomes exactly \(dt\).

Step 1: Transform the denominator. Using \[ (\sin x+\cos x)^2 = 1+\sin 2x, \] we get \[ \sin 2x=t^2-1. \] Hence, \[ 9+5\sin 2x = 9+5(t^2-1) = 4+5t^2. \] Therefore, \[ I = \int \frac{dt}{4+5t^2}. \]

Step 2: Change the limits. When \[ x=0, \] \[ t=\sin0+\cos0=1. \] When \[ x=\frac{\pi}{4}, \] \[ t=\frac{\sqrt2}{2}+\frac{\sqrt2}{2} =\sqrt2. \] Thus, \[ I = \int_{1}^{\sqrt2} \frac{dt}{4+5t^2}. \]

Step 3: Evaluate the integral. Using \[ \int\frac{dx}{a^2+b^2x^2} = \frac{1}{ab} \tan^{-1}\left(\frac{bx}{a}\right)+C, \] with \[ a=2, \qquad b=\sqrt5, \] we obtain \[ I = \frac{1}{2\sqrt5} \left[ \tan^{-1} \left( \frac{\sqrt5\,t}{2} \right) \right]_{1}^{\sqrt2}. \] \[ = \frac{1}{2\sqrt5} \left( \tan^{-1} \frac{\sqrt{10}}{2} - \tan^{-1} \frac{\sqrt5}{2} \right). \] Therefore, \[ \boxed{ \int_{0}^{\pi/4} \frac{\cos x-\sin x}{9+5\sin 2x}\,dx = \frac{1}{2\sqrt5} \left( \tan^{-1}\frac{\sqrt{10}}{2} - \tan^{-1}\frac{\sqrt5}{2} \right) } \] \[ \boxed{\text{Answer = (B)}} \]