Evaluating the Definite Integral
The problem asks us to find the value of the definite integral:
\(I = \int_0^{\frac{\pi}{2}} \frac{a + \sin x}{2a + \sin x + \cos x} dx\)
We can solve this integral using a standard property of definite integrals. One such property is:
\(\int_0^a f(x) dx = \int_0^a f(a-x) dx\)
In our case, the upper limit is \(a = \frac{\pi}{2}\). Let's apply this property to our integral \(I\). We replace \(x\) with \((\frac{\pi}{2} - x)\):
\(\sin x \rightarrow \sin\left(\frac{\pi}{2} - x\right) = \cos x\)
\(\cos x \rightarrow \cos\left(\frac{\pi}{2} - x\right) = \sin x\)
So, the integral becomes:
\(I = \int_0^{\frac{\pi}{2}} \frac{a + \cos x}{2a + \cos x + \sin x} dx\)
Let's call this transformed integral \(I'\). So, we have:
\(I' = \int_0^{\frac{\pi}{2}} \frac{a + \cos x}{2a + \sin x + \cos x} dx\)
Combining Integrals
Now, let's add the original integral \(I\) and the transformed integral \(I'\):
\(I + I' = \int_0^{\frac{\pi}{2}} \frac{a + \sin x}{2a + \sin x + \cos x} dx + \int_0^{\frac{\pi}{2}} \frac{a + \cos x}{2a + \sin x + \cos x} dx\)
Since the denominators are the same, we can combine the numerators:
\(I + I' = \int_0^{\frac{\pi}{2}} \frac{(a + \sin x) + (a + \cos x)}{2a + \sin x + \cos x} dx\)
\(I + I' = \int_0^{\frac{\pi}{2}} \frac{2a + \sin x + \cos x}{2a + \sin x + \cos x} dx\)
The integrand simplifies to 1:
\(I + I' = \int_0^{\frac{\pi}{2}} 1 dx\)
Now, we evaluate this simple integral:
\(I + I' = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2}\)
So, we have our first key equation:
\(I + I' = \frac{\pi}{2} \quad (1)\)
Considering the Difference of Integrals
Let's also consider the difference between the original integral \(I\) and the transformed integral \(I'\):
\(I - I' = \int_0^{\frac{\pi}{2}} \frac{a + \sin x}{2a + \sin x + \cos x} dx - \int_0^{\frac{\pi}{2}} \frac{a + \cos x}{2a + \sin x + \cos x} dx\)
Combine the numerators:
\(I - I' = \int_0^{\frac{\pi}{2}} \frac{(a + \sin x) - (a + \cos x)}{2a + \sin x + \cos x} dx\)
\(I - I' = \int_0^{\frac{\pi}{2}} \frac{\sin x - \cos x}{2a + \sin x + \cos x} dx\)
To evaluate this integral, let \(u = 2a + \sin x + \cos x\). Then, the derivative is \(du = (\cos x - \sin x) dx\). This means \((\sin x - \cos x) dx = -du\).
The integral becomes:
\(\int \frac{-du}{u} = -\ln|u| = -\ln|2a + \sin x + \cos x|\)
Now, evaluate this definite integral from \(0\) to \(\frac{\pi}{2}\):
\(I - I' = [-\ln|2a + \sin x + \cos x|]_0^{\frac{\pi}{2}}\)
\(I - I' = \left( -\ln\left|2a + \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right)\right| \right) - \left( -\ln\left|2a + \sin(0) + \cos(0)\right| \right)\)
\(I - I' = \left( -\ln|2a + 1 + 0| \right) - \left( -\ln|2a + 0 + 1| \right)\)
\(I - I' = -\ln(2a+1) + \ln(2a+1) = 0\)
(Assuming \(2a+1 > 0\), which is usually the case in such problems).
So, we have our second key equation:
\(I - I' = 0 \quad (2)\)
From equation (2), we get \(I = I'\).
Final Calculation
Substitute \(I' = I\) into equation (1):
\(I + I = \frac{\pi}{2}\)
\(2I = \frac{\pi}{2}\)
\(I = \frac{\pi}{4}\)
Therefore, the value of the definite integral is \(\frac{\pi}{4}\).
Similar Questions
What is the value of \(8 I_1^2\)
What is the value of I2 ?
What is \(\mathop \smallint \nolimits_{{{\rm{e}}^{ - 1}}}^{{{\rm{e}}^2}} \left| {\frac{{\ln {\rm{x}}}}{{\rm{x}}}} \right|{\rm{dx}}\) equal to?
What is \(\mathop \smallint \limits_{ - 2}^2 {\rm{x\;dx}} - \mathop \smallint \limits_{ - 2}^2 \left[ {\rm{x}} \right]{\rm{dx}}\) equal to, where [⋅] is the greatest integer function?
- \(\mathop \smallint \limits_0^{\frac{\pi }{2}} {e^{\sin x}}\cos x\;dx\) is equal to
What is \(\displaystyle\int_0^1 \ln \left(\frac{1}{x}−1\right)\) dx equal to ?
- If \(\displaystyle\int_0^{\pi/2}\) (sin 4 x + cos 4 x)dx = k, then what is the value of \(\displaystyle\int_0^{20 \pi}\) (sin 4 x + cos 4 x)dx ?
- What is \(\displaystyle\int_{−\pi/2}^{\pi/2}\) (e cos x sin x + e sin x cos x)dx equal to ?
What is \(\rm \int^\pi _0 ln\left(tan\frac{x}{2}\right) dx\) equal to?
What is the area bounded by y = [x], where [⋅] is the greatest integer function, the x-axis and the lines x = -1.5 and x = -1.8?
Important Questions from Definite Integrals
The Legendre polynomials P n(x), n = 0, 1, 2, ..., satisfying the orthogonailty condition \(\int_{{\rm{ - 1}}}^{\rm{1}} {{{\rm{P}}_{\rm{n}}}\left( {\rm{x}} \right){{\rm{P}}_{\rm{m}}}} \left( {\rm{x}} \right){\rm{dx}}\,{\rm{ = }}\,\frac{{\rm{2}}}{{{\rm{2n + 1}}}}{{\rm{\delta }}_{{\rm{nm}}}}\) on the interval [-1, +1], may be defined by the Rodrigues formula P n(x) = \(\frac{{\rm{1}}}{{{{\rm{2}}^{\rm{n}}}{\rm{n!}}}}\frac{{{{\rm{d}}^{\rm{n}}}}}{{{\rm{d}}{{\rm{x}}^{\rm{n}}}}}{\left( {{{\rm{x}}^{\rm{2}}}{\rm{ - 1}}} \right)^{\rm{n}}}\) . The value of the definite integral \(\int_{{\rm{ - 1}}}^{\rm{1}} {\left( {{\rm{4 + 2x - 3}}{{\rm{x}}^{\rm{2}}}{\rm{ + 4}}{{\rm{x}}^{\rm{3}}}} \right){{\rm{P}}_{\rm{3}}}\left( {\rm{x}} \right){\rm{dx}}} \) is
- \(\rm \displaystyle\int_1^3 (e^{\log x} + 1) dx\) is equal to
A parametric curve is defined \(x = cos\left(\frac{\Pi t}{2}\right) , Y= sin\left(\frac{\Pi t}{2}\right)\) in the range of \(0\leq t\leq 1\) . It is rotated about X-axis by 360°.
‘What is the area of the surface generated?
if \(\displaystyle\int\dfrac{\sin x}{\sin (x-a)}dx=Ax+B\log |sin(x-a)|+ C\) where A, B and c are real constants then:
Which of the following is NOT a property of definite integral?
