# Definite Integration Set 4

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Example -7

Prove that \begin{align}I = \int\limits_0^{2\pi } {\frac{{x\;{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx = {\pi ^2}.}\end{align}

Solution: Applying property -9 on I and then adding the original and modified forms of I, we obtain:

\begin{align}2I = 2\pi \int\limits_0^{2\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} \end{align}

The function to be integrated is symmetric about $$\pi$$ (verify this by the substitution $$x \to 2\pi --x$$).

Therefore:

$2I = 4\pi \int\limits_0^\pi {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx}$

The function to be integrated is still symmetric, about \begin{align}x = \frac{\pi }{2};\end{align} verify this by the substitution $$x \to \pi –x.$$

Therefore:

$2I = 8\pi \int\limits_0^{\pi {\rm{/2}}} {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

Now we apply property -9 on the integral above, i.e., we use$$x \to \frac{\pi }{2}--x.$$Thus,

$2I = 8\pi \int\limits_0^{\pi {\rm{/2}}} {\frac{{{{\cos }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

Adding (1) and (2), we obtain

\begin{align}&4I = 8\pi \int\limits_0^{\pi {\rm{/2}}} {\frac{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} \\\,\,\,\,\,\, &\;\;\;= 8\pi \int\limits_0^{\pi {\rm{/2}}} {dx} \\\,\,\,\,\,\, &\;\;\;= 4{\pi ^2}\\ \Rightarrow\quad I &\;\;\;= {\pi ^2}\end{align}

Example -8

Find a lower bound and an upper bound for the integral

$I = \int\limits_0^1 {\frac{1}{{\sqrt {1 + {x^6}} }}dx}$

Solution: By this question, we mean that we need to find two values m and M between which I lies, i.e.,

$m < I < M.$

Technically, any number less than I, say –(1 billion) is a valid lower bound for I and any number greater than I, say (1 billion), would be a valid upper bound. But how useful would it be to state the (trivial) fact that I lies between (– 1 billion) and (1 billion)? Not much.

We want ‘tight’ bounds, i.e., narrow ranges in which I could lie. Thus, Mm should be as small as possible so that we have an accurate idea about the approximate value of I.

Lets obtain an upper bound first:

We have,

\begin{align}&1 + {x^6} > 1 \quad\;\;\; {\rm{for}} \quad\;\;\; x \in (0,1)\\ \Rightarrow \quad &\sqrt {1 + {x^6}} > 1 \quad{\rm{for}} \quad x \in (0,1)\\ \Rightarrow \quad &\frac{1}{{\sqrt {1 + {x^6}} }} < 1 \quad{\rm{for}} \quad x \in (0,1)\\ \Rightarrow \quad &\int\limits_0^1 {\frac{1}{{\sqrt {1 + {x^6}} }}} dx < \int\limits_0^1 {1 \cdot dx} & \\ \Rightarrow\quad &I < 1\end{align}

Thus we now know that I is less than 1.

Lets obtain a ‘good’ lower bound now:

Since $$x \in (0,1),\,\,\,\,\,\,{x^6} < x.$$ Thus,

\begin{align}& \qquad \quad\;\; 1 + {x^6} < 1 + x < {(1 + x)^2} \quad \qquad \qquad {\rm{for}} \quad x \in (0,1)\\&\Rightarrow \qquad \sqrt {1 + {x^6}} < 1 + x \quad \qquad \qquad \quad \qquad {\rm\;\;\;{for}} \quad x \in (0,1)\\& \Rightarrow \qquad \int\limits_0^1 {\frac{1}{{\sqrt {1 + {x^6}} }}dx} > \int\limits_0^1 {\frac{1}{{1 + x}}} dx\\ &\Rightarrow\qquad I > \ln 2\end{align}

Thus, we now have a fair idea about the approximate value of I:

$\ln 2 < I < 1$

Try to obtain tighter bounds for yourself.

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