Question

The value of $\int_{ - \dfrac{{3\pi }}{2}}^{ - \dfrac{\pi }{2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]dx} $ is equal to
A. $\dfrac{{{\pi ^4}}}{{32}} + \dfrac{\pi }{2}$
B. $\dfrac{\pi }{2}$
C. $\dfrac{\pi }{4} - 1$
D. $\dfrac{{{\pi ^4}}}{{32}}$

Answer 1

Hint: First assume $x + \pi $. Then change the limit according to it. After that break the integration into two parts and apply cosine property i.e., $\cos \left( {x + 2\pi } \right) = \cos x$. After that integrate the function and substitute the values to get the desired result.

Complete step by step answer:
Let us assume $t = x + \pi $.
Then, differentiate the function to get the value of $dt$,
$ \Rightarrow dt = dx$
Now substitute the upper limit in the function,
$ \Rightarrow t = - \dfrac{\pi }{2} + \pi $
Take LCM on the right side,
$ \Rightarrow t = \dfrac{{ - \pi + 2\pi }}{2}$
Simplify the term,
$ \Rightarrow t = \dfrac{\pi }{2}$
So, the new upper limit is $\dfrac{\pi }{2}$.
Now substitute the lower limit in the function,
$ \Rightarrow t = - \dfrac{{3\pi }}{2} + \pi $
Take LCM on the right side,
$ \Rightarrow t = \dfrac{{ - 3\pi + 2\pi }}{2}$
Simplify the term,
$ \Rightarrow t = - \dfrac{\pi }{2}$
So, the new lower limit is $ - \dfrac{\pi }{2}$.
The new equation will be,
$ \Rightarrow \int_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\left[ {{t^3} + {{\cos }^2}\left( {t + 2\pi } \right)} \right]dt} $
Break the terms into two parts,
\[ \Rightarrow \int_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {{t^3}dt} + \int_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {{{\cos }^2}\left( {t + 2\pi } \right)dt} \]
Now, we know that,
$\cos \left( {x + 2\pi } \right) = \cos x$
Using this property, the above integral value will be,
\[ \Rightarrow \int_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {{t^3}dt} + \int_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {{{\cos }^2}tdt} \]
We know that,
${\cos ^2}x = \dfrac{{1 + \cos x}}{2}$
Substitute the value in the above equation,
\[ \Rightarrow \int_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {{t^3}dt} + \int_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{1 + \cos 2t}}{2}dt} \]
Now integrate the terms,
\[ \Rightarrow \left[ {\dfrac{{{t^4}}}{4}} \right]_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} + \dfrac{1}{2}\left[ {t + \dfrac{{\sin 2t}}{2}} \right]_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}}\]
Substitute the limits,
\[ \Rightarrow \left[ {\dfrac{{{{\left( {\dfrac{\pi }{2}} \right)}^4} - {{\left( { - \dfrac{\pi }{2}} \right)}^4}}}{4}} \right] + \dfrac{1}{2}\left[ {\left( {\dfrac{\pi }{2} + \dfrac{{\sin 2 \times \dfrac{\pi }{2}}}{2}} \right) - \left( { - \dfrac{\pi }{2} + \dfrac{{\sin 2 \times - \dfrac{\pi }{2}}}{2}} \right)} \right]\]
Simplify the terms,
\[ \Rightarrow \left[ {\dfrac{{\dfrac{{{\pi ^4}}}{{16}} - \dfrac{{{\pi ^4}}}{{16}}}}{4}} \right] + \dfrac{1}{2}\left[ {\dfrac{\pi }{2} + \dfrac{{\sin \pi }}{2} + \dfrac{\pi }{2} + \dfrac{{\sin \pi }}{2}} \right]\]
Again, simplify the terms,
\[ \Rightarrow 0 + \dfrac{1}{2}\left[ {\dfrac{\pi }{2} + 0 + \dfrac{\pi }{2} + 0} \right]\]
Add the terms in the bracket,
\[ \Rightarrow \dfrac{1}{2} \times \pi \]
Multiply the terms,
$\therefore \dfrac{\pi }{2}$
Thus, the value of $\int_{ - \dfrac{{3\pi }}{2}}^{ - \dfrac{\pi }{2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]dx} $ is $\dfrac{\pi }{2}$.

Hence, option (B) is correct.

Additional Information: Differentiation and integration are the two important concepts of calculus. Calculus is a branch of mathematics that deals with the study of problems involving a continuous change in the values of quantities. Differentiation refers to simplifying a complex function into simpler functions. Integration generally refers to summing up the smaller function to form a bigger unit.
Indefinite integrals are those integrals that do not have any limit of integration. It has an arbitrary constant. Definite integrals are those integrals which have an upper and lower limit. Definite integral has two different values for the upper limit and lower limit when they are evaluated. The final value of a definite integral is the value of integral to the upper limit minus the value of the definite integral for the lower limit.

Note: To solve these types of questions one should know the basic concepts of integral calculus. Students might make mistakes in assuming the value $t = x + \pi $. They might take $t = x + 3\pi $, which will lead to lengthy calculations or even leads to wrong answers.