Definite Integration Set 7

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Example -13

Prove that if \(k \in {\mathbb{Z}^ + },\)

\[\frac{{\sin 2kx}}{{\sin x}} = 2\left[ {\cos x + \cos 3x + ........ + \cos (2k - 1)x} \right]\]

Hence or otherwise, prove that

\[\int\limits_{\rm{0}}^{\pi {\rm{/2}}} {\sin 2kx\cot xdx = \pi {\rm{/2}}} \]

Solution: Let us first directly try to prove the second part using the technique of recursion.

Let  \({I_k} = \int\limits_0^{\pi {\text{/2}}} {\sin 2kx\cot x\,dx} \)

\[\begin{align}& \Rightarrow\quad \,{I_1} = \int\limits_0^{\pi {\rm{/2}}} {\sin 2x\cot x\,dx} \\\,\,\,\,\,\, &\qquad\quad\;\;= 2\int\limits_0^{\pi {\rm{/2}}} {{{\cos }^2}x\,dx} \\\,\,\,\,\,\, &\qquad\quad\;\;= 2\int\limits_0^{\pi {\rm{/2}}} {\left[ {\frac{{1 + \cos 2x}}{2}} \right]dx} \\\,\,\,\,\,\, &\qquad\quad\;\;= \frac{\pi }{2}\end{align}\]

Also, \(\,\,\,{I_{k + 1}} - {I_k} = \int\limits_0^{\pi {\rm{/2}}} {\left( {\sin (2k + 2)x - \sin 2kx} \right)\cot x\,dx} \)

\[\begin{align}&\,\,\,\,\, = 2\int\limits_0^{\pi {\rm{/2}}} {\cos (2k + 1)x\sin x \cdot \cot x\,dx} \\\,\,\,\,\, &= \int\limits_0^{\pi {\rm{/2}}} {\left( {2\cos x \cdot \cos (2k + 1)x} \right)dx} \\\,\,\,\,\, &= \int\limits_0^{\pi {\rm{/2}}} {\left( {\cos (2k + 2)x + \cos 2kx} \right)dx} \\\,\,\,\,\, &= \left. {\frac{{\sin (2k + 2)x}}{{2k + 2}}} \right|_0^{\pi {\rm{/2}}} - \left. {\frac{{\sin 2kx}}{{2k}}} \right|_0^{\pi {\rm{/2}}}\\\,\,\,\,\, &= 0 - 0\,\,\,\,\,\\\,\,\,\,\, &= 0\end{align}\]

Thus,

\({I_k} = \frac{\pi }{2}\) for all \(k \in {\mathbb{Z}^ + }\)

Now we’ll use the first result mentioned in the question to prove the second part.

The proof of the first result is simple:

\[\begin{array}{l}2\sin x\left[ {\cos x + \cos 3x + .......... + \cos (2k - 1)x} \right]\\\,\,\,\, = \sin 2x + (\sin 4x - \sin 2x) + ..... + (\sin 2kx - \sin (2k - 2)x)\\\,\,\,\, = \sin 2kx\end{array}\]

Thus, the stated assertion is valid

Now,

\[\begin{align}&I = \int\limits_0^{\pi {\rm{/2}}} {\sin 2kx\cot x\,dx} \\\,\,\, &\;= 2\int\limits_0^{\pi {\rm{/2}}} {\cos x\left[ {\cos x + \cos 3x + .......\cos (2k - 1)x} \right]dx} & \left\{ \begin{array}{l}{\rm{Using \;the }}\\{\rm{first\; result}}\end{array} \right\}\\\,\,\, &\;= \int\limits_0^{\pi {\rm{/2}}} {\left\{ {1 + \cos\ 2x + (\cos 4x - \cos 2x) + ...... + \cos 2kx - \cos (2k - 2)x} \right\}dx} \\\,\,\, &\;= \int\limits_0^{\pi {\rm{/2}}} {(1 + \cos 2kx)dx} \\\,\,\, &\;= \frac{\pi }{2} & \left\{ {\cos \;2kx\,\,{\rm{integrates\; to\; }}0} \right\}\\ \end{align}\]

Example -14

Suppose that \(g(x)\) is an even function, and \(f(x) = \int\limits_0^x {g(t)dt.} \)

Is \(f(x)\) even or odd, or neither?

Solution: \(f( - x) = \int\limits_0^{ - x} {g(t)dt} \)

If we let \(t = - y,\) the limits of integration change from (0 to –x) to (0 to x). Thus,

\[\begin{align}& f( - x) = \int\limits_0^x {g( - y)( - dy)} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad= - \int\limits_0^x {g(y)dy\;\;\;\;\; (\because \,\,\,g{\text{ is even)}}} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad= - f(x) \\ \end{align} \]

Thus, \(f(x)\) is odd.

Example -15

Let x > 0. Evaluate \(\int\limits_0^x {[t]dt.} \)

Solution: The function to be integrated is discontinuous at all integer points. Therefore, we integrate it piecewise:

\[\begin{align}&\int\limits_0^x {[t]dt} = \int\limits_0^1 {[t]dt} + \int\limits_1^2 {[t]dt} + .....\int\limits_{[x] - 1}^{[x]} {[t]dt} + \int\limits_{[x]}^x {[t]dt} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad= 0 \cdot \int\limits_0^1 {dt} + 1 \cdot \int\limits_1^2 {dt} + 2 \cdot \int\limits_2^3 {dt} + ..... + \left( {[x] - 1} \right) \cdot \int\limits_{[x] - 1}^{[x]} {dt} + \int\limits_{[x]}^x {[t]dt} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad= 0 + 1 + 2 + .......\left( {[x] - 1} \right) + \left( {x - [x]} \right)\left( {[x]} \right) & \left( \begin{array}{l}{\rm{Verify\; that\; the\; last}}\\{\rm{term\; is\; correct}}\end{array} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad= \frac{{[x]\left( {[x] - 1} \right)}}{2} + [x]\left( {x - [x]} \right)\end{align}\]

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