The value of \(\mathop \smallint \limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} \frac{{dx}}{{{{\sin }^2}x{{\cos }^2}x}}\) is
This problem asks us to find the value of a definite integral involving trigonometric functions. The integral is given by:
\(\mathop \smallint \limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} \frac{{dx}}{{{{\sin }^2}x{{\cos }^2}x}}\)
To solve this definite integral, we first need to find the antiderivative of the integrand \(\frac{1}{{\sin^2 x \cos^2 x}}\).
Simplifying the Integrand for Definite Integral Evaluation
The integrand can be simplified using the identity \(\sin^2 x + \cos^2 x = 1\). We can rewrite the expression as:
\(\frac{1}{{\sin^2 x \cos^2 x}} = \frac{{\sin^2 x + \cos^2 x}}{{\sin^2 x \cos^2 x}}\)
Now, we can split this fraction into two terms:
\(\frac{{\sin^2 x}}{{\sin^2 x \cos^2 x}} + \frac{{\cos^2 x}}{{\sin^2 x \cos^2 x}} = \frac{1}{{\cos^2 x}} + \frac{1}{{\sin^2 x}}\)
Using reciprocal identities, we know that \(\frac{1}{{\cos^2 x}} = \sec^2 x\) and \(\frac{1}{{\sin^2 x}} = \csc^2 x\). So, the integrand simplifies to:
\(\sec^2 x + \csc^2 x\)
This simplification makes finding the antiderivative much easier for evaluating the definite integral.
Finding the Antiderivative
Now we need to find the indefinite integral of the simplified expression:
\(\int (\sec^2 x + \csc^2 x) dx\)
We know the standard integrals for \(\sec^2 x\) and \(\csc^2 x\):
- \(\int \sec^2 x dx = \tan x + C_1\)
- \(\int \csc^2 x dx = -\cot x + C_2\)
Combining these, the antiderivative of \(\sec^2 x + \csc^2 x\) is \(\tan x - \cot x\). We don't need the constant of integration \(C\) for definite integrals.
Evaluating the Definite Integral with Integration Limits
Now we evaluate the definite integral using the fundamental theorem of calculus. The limits of integration are from \(\frac{\pi}{4}\) to \(\frac{\pi}{3}\).
\(\mathop \smallint \limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} (\sec^2 x + \csc^2 x) dx = [\tan x - \cot x]_{\frac{\pi}{4}}^{\frac{\pi}{3}}\)
This means we need to calculate \((\tan(\frac{\pi}{3}) - \cot(\frac{\pi}{3})) - (\tan(\frac{\pi}{4}) - \cot(\frac{\pi}{4}))\).
Calculating Values at the Limits
- For the upper limit \(\frac{\pi}{3}\):
- \(\tan(\frac{\pi}{3}) = \sqrt{3}\)
- \(\cot(\frac{\pi}{3}) = \frac{1}{\sqrt{3}}\)
- For the lower limit \(\frac{\pi}{4}\):
- \(\tan(\frac{\pi}{4}) = 1\)
- \(\cot(\frac{\pi}{4}) = 1\)
Final Calculation of the Definite Integral
Substitute these values back into the expression:
\([\tan x - \cot x]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) - (1 - 1)\)
\(= \left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) - 0\)
\(= \sqrt{3} - \frac{1}{\sqrt{3}}\)
To simplify \(\sqrt{3} - \frac{1}{\sqrt{3}}\), find a common denominator:
\(= \frac{\sqrt{3} \cdot \sqrt{3}}{1 \cdot \sqrt{3}} - \frac{1}{\sqrt{3}}\)
\(= \frac{3}{\sqrt{3}} - \frac{1}{\sqrt{3}}\)
\(= \frac{3 - 1}{\sqrt{3}}\)
\(= \frac{2}{\sqrt{3}}\)
Thus, the value of the definite integral is \(\frac{2}{\sqrt{3}}\). This calculus problem requires understanding trigonometric functions and their integrals.
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
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:
The value of integral \(x\over cos^2 x\) is equal to
- \(\mathop \smallint \limits_{ - 1}^2 x\left| x \right|\;dx\) is equal to
