Question

The value of the integral $\int_0^{\pi/2} \frac{dx}{(1 + \sin^2 x)}$ lies in smallest interval out of given options is:

• A. $\left(\frac{\pi}{2}, \pi\right)$
• B. $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
• C. $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$
• D. $\left(\frac{\pi}{8}, \frac{\pi}{2}\right)$

Answer 1

Answer: (C)

(C)

Answer 2

To evaluate the integral $I = \int_0^{\pi/2} \frac{dx}{(1 + \sin^2 x)}$, we follow these steps:

1. Transform the integrand: Divide the numerator and denominator by $\cos^2 x$: $I = \int_0^{\pi/2} \frac{1/\cos^2 x}{(1 + \sin^2 x)/\cos^2 x} dx$

   $I = \int_0^{\pi/2} \frac{\sec^2 x}{\sec^2 x + \tan^2 x} dx$

   Substitute $\sec^2 x = 1 + \tan^2 x$ in the denominator:

   $I = \int_0^{\pi/2} \frac{\sec^2 x}{1 + \tan^2 x + \tan^2 x} dx$

   $I = \int_0^{\pi/2} \frac{\sec^2 x}{1 + 2\tan^2 x} dx$

2. Substitute and change limits: Let $t = \tan x$. Then $dt = \sec^2 x dx$.

   When $x = 0$, $t = \tan 0 = 0$.

   When $x = \pi/2$, $t = \tan(\pi/2) \to \infty$.

   The integral becomes:

   $I = \int_0^{\infty} \frac{dt}{1 + 2t^2}$

3. Evaluate the transformed integral:

   Factor out 2 from the denominator:

   $I = \frac{1}{2} \int_0^{\infty} \frac{dt}{\frac{1}{2} + t^2}$

   Rewrite $\frac{1}{2}$ as $\left(\frac{1}{\sqrt{2}}\right)^2$:

   $I = \frac{1}{2} \int_0^{\infty} \frac{dt}{\left(\frac{1}{\sqrt{2}}\right)^2 + t^2}$

   This is a standard integral of the form $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \arctan\left(\frac{x}{a}\right)$. Here $a = \frac{1}{\sqrt{2}}$.

   $I = \frac{1}{2} \left[ \frac{1}{1/\sqrt{2}} \arctan\left(\frac{t}{1/\sqrt{2}}\right) \right]_0^{\infty}$

   $I = \frac{1}{2} \left[ \sqrt{2} \arctan(\sqrt{2}t) \right]_0^{\infty}$

   $I = \frac{\sqrt{2}}{2} \left[ \arctan(\sqrt{2}t) \right]_0^{\infty}$

   $I = \frac{1}{\sqrt{2}} \left[ \arctan(\infty) - \arctan(0) \right]$

   $I = \frac{1}{\sqrt{2}} \left[ \frac{\pi}{2} - 0 \right]$

   $I = \frac{\pi}{2\sqrt{2}}$

4. Simplify and compare with options:

   The value of the integral is $I = \frac{\pi}{2\sqrt{2}} = \frac{\pi\sqrt{2}}{4}$.

   Now we need to find which of the given intervals contains this value and is the smallest.

   Let's compare $I$ with the bounds of the given intervals:

   $I = \frac{\pi\sqrt{2}}{4} \approx \frac{3.14159 \times 1.4142}{4} \approx \frac{4.4428}{4} \approx 1.1107$.

   The options are:

   (A) $\left(\frac{\pi}{2}, \pi\right) \approx (1.5708, 3.1416)$

   (B) $\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \approx (0.7854, 1.5708)$

   (C) $\left(\frac{\pi}{3}, \frac{\pi}{2}\right) \approx (1.0472, 1.5708)$

   (D) $\left(\frac{\pi}{8}, \frac{\pi}{2}\right) \approx (0.3927, 1.5708)$

   We need to check if $I$ lies in these intervals more precisely.

   Is $I < \frac{\pi}{2}$?

   $\frac{\pi}{2\sqrt{2}} < \frac{\pi}{2} \implies \frac{1}{\sqrt{2}} < 1 \implies 1 < \sqrt{2}$, which is true. So $I$ is less than $\pi/2$.

   Is $I > \frac{\pi}{3}$?

   $\frac{\pi}{2\sqrt{2}} > \frac{\pi}{3} \implies \frac{1}{2\sqrt{2}} > \frac{1}{3} \implies 3 > 2\sqrt{2}$. Squaring both sides: $9 > 8$, which is true. So $I > \pi/3$.

   Therefore, $I \in \left(\frac{\pi}{3}, \frac{\pi}{2}\right)$. This means option (C) contains $I$.

   Since $\pi/3 > \pi/4 > \pi/8$, if $I \in (\pi/3, \pi/2)$, then $I$ must also be in $(\pi/4, \pi/2)$ and $(\pi/8, \pi/2)$.

   So, options (B), (C), and (D) all contain the value of the integral.

   We need to find the smallest interval among the given options that contains $I$. The "smallest" interval refers to the one with the shortest length.

   Lengths of the intervals:

   Length of (B) = $\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$

   Length of (C) = $\frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$

   Length of (D) = $\frac{\pi}{2} - \frac{\pi}{8} = \frac{3\pi}{8}$

   Comparing the lengths: $\frac{\pi}{6} < \frac{\pi}{4} < \frac{3\pi}{8}$ (since $4/24 < 6/24 < 9/24$).

   The smallest length is $\frac{\pi}{6}$, which corresponds to interval (C).