Question

Evaluate the definite integral: \[ \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{x+\frac{\pi}{2}} {2-\sin^2 x} \,dx \]

• A. \[ \frac{\pi^2}{6\sqrt3} \]
• B. \[ \frac{\pi}{\sqrt2} \tan^{-1} \left( \frac{\sqrt3}{\sqrt2} \right) \]
• C. \[ \frac{\pi^2}{3\sqrt2} \]
• D. \[ \frac{\pi}{2} \tan^{-1} \left( \frac{\sqrt3}{2} \right) \]

Hint:

Whenever a definite integral has symmetric limits \([-a,a]\), first test whether the integrand is odd or even. This often eliminates a large portion of the calculation and converts the problem into a much simpler integral.

Answer 1

The Correct Option is B

Concept: Whenever a definite integral is evaluated over a symmetric interval of the form \[ [-a,a], \] it is advantageous to separate the integrand into its odd and even parts. The integral of an odd function over symmetric limits is always zero, while the integral of an even function over symmetric limits can be simplified using \[ \int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx. \] After applying symmetry, the remaining integral can be evaluated using an appropriate trigonometric substitution.

Step 1: Split the integral into two separate parts. Let \[ I= \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{x+\frac{\pi}{2}} {2-\sin^2 x} \,dx. \] Separating the numerator, \[ I = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{x} {2-\sin^2 x} \,dx + \frac{\pi}{2} \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{1} {2-\sin^2 x} \,dx. \] Let \[ I=I_1+I_2, \] where \[ I_1= \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{x} {2-\sin^2 x} \,dx \] and \[ I_2= \frac{\pi}{2} \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{1} {2-\sin^2 x} \,dx. \]

Step 2: Evaluate \(I_1\) using symmetry. Consider \[ f(x)= \frac{x} {2-\sin^2 x}. \] Now, \[ f(-x) = \frac{-x} {2-\sin^2(-x)} = \frac{-x} {2-\sin^2 x} = -f(x). \] Therefore \(f(x)\) is an odd function. Since the limits are symmetric, \[ I_1 = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} f(x)\,dx = 0. \]

Step 3: Simplify \(I_2\). The integrand \[ g(x)=\frac{1}{2-\sin^2 x} \] is an even function because \[ g(-x) = g(x). \] Hence \[ I_2 = \frac{\pi}{2} \cdot 2 \int_0^{\frac{\pi}{3}} \frac{1} {2-\sin^2 x} \,dx. \] Thus, \[ I_2 = \pi \int_0^{\frac{\pi}{3}} \frac{1} {2-\sin^2 x} \,dx. \]

Step 4: Convert the integral into a standard form. Using \[ \sec^2x = 1+\tan^2x, \] divide numerator and denominator by \(\cos^2x\): \[ I_2 = \pi \int_0^{\frac{\pi}{3}} \frac{\sec^2x} {2\sec^2x-\tan^2x} \,dx. \] Since \[ 2\sec^2x-\tan^2x = 2(1+\tan^2x)-\tan^2x = 2+\tan^2x, \] we get \[ I_2 = \pi \int_0^{\frac{\pi}{3}} \frac{\sec^2x} {2+\tan^2x} \,dx. \]

Step 5: Apply substitution. Let \[ t=\tan x. \] Then \[ dt=\sec^2x\,dx. \] Changing limits, \[ x=0 \Rightarrow t=0, \] \[ x=\frac{\pi}{3} \Rightarrow t=\sqrt3. \] Therefore, \[ I_2 = \pi \int_0^{\sqrt3} \frac{dt} {2+t^2}. \]

Step 6: Integrate using the standard formula. Using \[ \int \frac{dx}{a^2+x^2} = \frac1a \tan^{-1} \left( \frac{x}{a} \right), \] with \[ a=\sqrt2, \] we obtain \[ I_2 = \pi \left[ \frac1{\sqrt2} \tan^{-1} \left( \frac{t}{\sqrt2} \right) \right]_{0}^{\sqrt3}. \] Thus, \[ I_2 = \frac{\pi}{\sqrt2} \left[ \tan^{-1} \left( \frac{\sqrt3}{\sqrt2} \right) - \tan^{-1}(0) \right]. \] Since \[ \tan^{-1}(0)=0, \] we get \[ I_2 = \frac{\pi}{\sqrt2} \tan^{-1} \left( \frac{\sqrt3}{\sqrt2} \right). \]

Step 7: Compute the final value. Since \[ I_1=0, \] we obtain \[ I = I_1+I_2 = \frac{\pi}{\sqrt2} \tan^{-1} \left( \frac{\sqrt3}{\sqrt2} \right). \] Therefore, \[ \boxed{ \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{x+\frac{\pi}{2}} {2-\sin^2 x} \,dx = \frac{\pi}{\sqrt2} \tan^{-1} \left( \frac{\sqrt3}{\sqrt2} \right) } \] Hence the correct option is \[ \boxed{\text{(B)}}. \]