Question

Evaluate the indefinite integral as an infinite series. 43.\(\int {xcos({x^3})dx} \)

Short answer

\(\int {x\cos ({x^3})dx = \sum\limits_{n = 1}^\infty {{{( - 1)}^n}\frac{{{x^{6n + 2}}}}{{(6n + 2)(2n)!}} + C} } \)

Step-by-step solution

Step 1: Maclaurin series

We evaluate the given indefinite integral as an expression of an infinite series. Here we are going to use Maclaurin series representation of some functions and work our way through to obtain similar pattern of expression. Given indefinite integral

\(\cos (x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{(2n)!}}} \)

Step 2: Substitution

Recall that the Maclaurin series representation for \(\cos x\) is given by

\(\cos (x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{(2n)!}}} \)

It follows that

\(\begin{array}{l}\cos ({x^3}) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{{({x^3})}^{2n}}}}{{(2n)!}}} \\\cos ({x^3}) = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^{6n}}}}{{(2n)!}}} \end{array}\)

Step 3: Now multiply with x

The radius of convergence will remain the same, so \(R = \infty \). Now, all we need to do is to multiply it by by x – since this is a constant in regards to n:

\(\begin{array}{l}x\cos ({x^3}) = x\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{{({x^3})}^{2n}}}}{{(2n)!}}} \\\cos ({x^3}) = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^{6n + 1}}}}{{(2n)!}}} \end{array}\)

Step 4: Now perform integration

\(\begin{array}{l}\int {x\cos ({x^3})dx} = \int {\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^{6n + 1}}}}{{(2n)!}}} dx} \\ = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}}}{{(2n)!}}\int {{x^{6n + 1}}dx} } \end{array}\)

\(\begin{array}{l} = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}}}{{(2n)!}}\left( {\frac{{{x^{6n + 1}}}}{{6n + 2}}} \right) + C} \\ = \sum\limits_{n = 1}^\infty {{{( - 1)}^n}\frac{{{x^{6n + 1}}}}{{(2n)!(6n + 2)}} + C} \end{array}\)