The integral \(\int_0^{\frac{\pi}{2}} \frac{1}{3 + 2\sin x + \cos x} \, dx\) is equal to
\(\tan^{-1}(2)\)
\(\tan^{-1}(2) - \frac{\pi}{4}\)
\(\frac{1}{2}\tan^{-1}(2) - \frac{\pi}{8}\)
\(\frac{1}{2}\)
The Correct Option is B
Approach Solution - 1
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 \( \tan^{-1}(2) - \frac{\pi}{4} \).
Approach Solution -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): \(\tan^{-1}(2) - \frac{\pi}{4}\)
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Concepts Used:
Methods of Integration
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
Integration by Parts:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
Method of Integration Using Partial Fractions:
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Integration by Substitution Method
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
Integration by Decomposition
Reverse Chain Rule
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C