Question

# Let the integral $I = \int\limits_0^1 {\dfrac{{{x^3}\cos 3x}}{{2 + {x^2}}}dx}$. Then which condition on $I$ holds true$\left( a \right){\text{ }}\dfrac{{ - 1}}{2} < I < \dfrac{1}{2} \\ \left( b \right){\text{ }}\dfrac{{ - 1}}{3} < I < \dfrac{1}{3} \\ \left( c \right){\text{ - 1}} < I < 1 \\ \left( d \right){\text{ }}\dfrac{{ - 3}}{2} < I < \dfrac{3}{2} \\$

Hint: In this question we have to find the range of the integral $I$ and not the exact value so direct solving of integral using definite integral property won’t help, we have to think in such a way that what is the range of the trigonometric entity defined inside the integral and then form the integral’s range using this concept.

Given integral
$I = \int\limits_0^1 {\dfrac{{{x^3}\cos 3x}}{{2 + {x^2}}}dx}$………… (1)
Now as we know that cos x always lies between -1 to 1 therefore cos 3x also lies between -1 to 1.
$\Rightarrow - 1 \leqslant \cos 3x \leqslant 1$
Now multiply by ${x^3}$ in above equation we have,
$\Rightarrow - {x^3} \leqslant {x^3}.\cos 3x \leqslant {x^3}$
Now divide by $\left( {2 + {x^2}} \right)$ in above equation we have,
$\Rightarrow - \dfrac{{{x^3}}}{{2 + {x^2}}} \leqslant \dfrac{{{x^3}.\cos 3x}}{{2 + {x^2}}} \leqslant \dfrac{{{x^3}}}{{2 + {x^2}}}$
Now apply integration from 0 to 1 we have,
$\Rightarrow - \int\limits_0^1 {\dfrac{{{x^3}}}{{2 + {x^2}}}dx} \leqslant \int\limits_0^1 {\dfrac{{{x^3}.\cos 3x}}{{2 + {x^2}}}dx} \leqslant \int\limits_0^1 {\dfrac{{{x^3}}}{{2 + {x^2}}}} dx$
Now from equation (1)
$I = \int\limits_0^1 {\dfrac{{{x^3}\cos 3x}}{{2 + {x^2}}}dx}$
Therefore above equation becomes
$\Rightarrow - \int\limits_0^1 {\dfrac{{{x^3}}}{{2 + {x^2}}}dx} \leqslant I \leqslant \int\limits_0^1 {\dfrac{{{x^3}}}{{2 + {x^2}}}} dx$
$\Rightarrow - \int\limits_0^1 {\dfrac{{x.{x^2}}}{{2 + {x^2}}}dx} \leqslant I \leqslant \int\limits_0^1 {\dfrac{{x.{x^2}}}{{2 + {x^2}}}} dx$………. (2)
Now let $\left( {2 + {x^2}} \right) = t$…………………. (3)
Differentiate this equation w.r.t. x we have
$\left( {0 + 2x} \right)dx = dt$
$\therefore xdx = \dfrac{{dt}}{2}$………. (4)
Now from equation (3)
If x = 0, $\Rightarrow t = 0 + 2 = 2$
And If x = 1, $\Rightarrow t = 1 + 2 = 3$
So the integration limit is changed from 2 to 3.
Now from equation (2), (3) and (4) we have
$\Rightarrow - \int\limits_2^3 {\dfrac{{t - 2}}{t}\left( {\dfrac{{dt}}{2}} \right)} \leqslant I \leqslant \int\limits_2^3 {\dfrac{{t - 2}}{t}} \left( {\dfrac{{dt}}{2}} \right)$
$\Rightarrow - \int\limits_2^3 {\left( {1 - \dfrac{2}{t}} \right)\left( {\dfrac{{dt}}{2}} \right)} \leqslant I \leqslant \int\limits_2^3 {\left( {1 - \dfrac{2}{t}} \right)} \left( {\dfrac{{dt}}{2}} \right)$
Now as we know integration of 1 is t w.r.t. t and integration of $\dfrac{2}{t}$ is 2logt so, apply this in above equation we have,
$\Rightarrow - \left[ {t - 2\log t} \right]_2^3 \leqslant I \leqslant \left[ {t - 2\log t} \right]_2^3$
Now apply integral limit we have
$\Rightarrow - \dfrac{1}{2}\left[ {3 - 2\log 3 - 2 + 2\log 2} \right] \leqslant I \leqslant \dfrac{1}{2}\left[ {3 - 2\log 3 - 2 + 2\log 2} \right]$
As we know $\left( {\log a - \log b = \log \dfrac{a}{b}} \right)$ so, use this property we have,
$\Rightarrow - \dfrac{1}{2}\left[ {1 - 2\log \dfrac{3}{2}} \right] \leqslant I \leqslant \dfrac{1}{2}\left[ {1 - 2\log \dfrac{3}{2}} \right]$
Now as we know$\left( {\log \dfrac{3}{2} = 0.1761} \right)$, so substitute this value we have,
$\Rightarrow - \dfrac{1}{2}\left[ {1 - 2 \times 0.1761} \right] \leqslant I \leqslant \dfrac{1}{2}\left[ {1 - 2 \times 0.1761} \right]$
Now on simplifying we have,
$\Rightarrow - 0.3239 \leqslant I \leqslant 0.3239$
As 0.3239 is approximately equal to$\dfrac{1}{3}$.
$\Rightarrow - \dfrac{1}{3} \leqslant I \leqslant \dfrac{1}{3}$
Hence, option (b) is correct.

Note: Whenever we face such types of problems the key concept is the basic understanding of the range of the trigonometric entities present inside the integral. Sometimes we also may need to simplify our integral so using proper substitution methods along with basic logarithm properties, we can easily evaluate the integral and reach the right answer.