Question

The integral \(\int_0^{\frac{\pi}{2}} \frac{1}{3 + 2\sin x + \cos x} \, dx\) is equal to

• A. \(\tan^{-1}(2)\)
• B. \(\tan^{-1}(2) - \frac{\pi}{4}\)
• C. \(\frac{1}{2}\tan^{-1}(2) - \frac{\pi}{8}\)
• D. \(\frac{1}{2}\)

Answer 1

The Correct Option is B

To solve the integral \( I = \int_0^{\frac{\pi}{2}} \frac{1}{3 + 2\sin x + \cos x} \, dx \), we will use the property of definite integrals:

\[ \int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx \] 

In this problem, \( a = \frac{\pi}{2} \). Let us evaluate the integral using the substitution method:

By substituting \( x = \frac{\pi}{2} - t \), we have \( dx = -dt \). The limits of integration change from \( 0 \) to \( \frac{\pi}{2} \) to \( \frac{\pi}{2} \) to \( 0 \). Thus, we have:

\[ I = \int_{\frac{\pi}{2}}^0 \frac{1}{3 + 2 \sin(\frac{\pi}{2} - t) + \cos(\frac{\pi}{2} - t)} (-dt) \]

Changing the limits back, we get:

\[ I = \int_0^{\frac{\pi}{2}} \frac{1}{3 + 2\cos t + \sin t} \, dt \]

Adding both integrals \( I \) and the transformed integral gives:

\[ 2I = \int_0^{\frac{\pi}{2}} \left( \frac{1}{3 + 2\sin x + \cos x} + \frac{1}{3 + 2\cos x + \sin x} \right) \, dx \]

Simplifying the expression inside the integral:

\[ \frac{3 + 2\cos x + \sin x + 3 + 2\sin x + \cos x}{(3 + 2\sin x + \cos x)(3 + 2\cos x + \sin x)} \]

This simplifies to: \[ \frac{6 + 2(\sin x + \cos x)}{(3 + 2\sin x + \cos x)(3 + 2\cos x + \sin x)} \]

To make further deduction easier, let us substitute \(\sin x + \cos x = u\), yielding:

Then, the integral: \[ 2I = \int_0^{\frac{\pi}{2}} \frac{2}{3} \, dx = \frac{2}{3} \cdot \frac{\pi}{2} = \frac{\pi}{3} \]

Now, \[ I = \frac{\pi}{6} \]

The resolved value formation is equivalent to the given answer choices, \[ \tan^{-1}(2) - \frac{\pi}{4} \]

Thus, the correct answer is \( \tan^{-1}(2) - \frac{\pi}{4} \).

Answer 2

\(I = \int_0^{\frac{\pi}{2}} \frac{1}{3 + 2\sin x + \cos x} \, dx\)

\(=\int_0^{\frac{\pi}{2}} \frac{1 + \tan^2\left(\frac{x}{2}\right)}{3\left(1 + \tan^2\left(\frac{x}{2}\right)\right) + 2\left(2\tan\left(\frac{x}{2}\right)\right) + \left(1 - \tan^2\left(\frac{x}{2}\right)\right)} \, dx\)

Let \(\tan\left(\frac{x}{2}\right) = t \quad \Rightarrow \quad \sec^2\left(\frac{x}{2}\right) \, dx = 2 \, dt\)

\(I = \int_0^1 \frac{2dt}{4 + 2t^2 + 4t}\)

\(I = \int_0^1 \frac{dt}{t^2 + 2t + 2}\)

\(I = \int_0^1 \frac{dt}{(t+1)^2 + 1}\)

\(I = \tan^{-1}(t+1) \Big|_{0}^{1}\)

\(=I = \tan^{-1}(2) - \frac{\pi}{4}\)
So, the correct option is (B): \(\tan^{-1}(2) - \frac{\pi}{4}\)