Question

PROHIBITED
The coefficient of x8 in the expansion of sin(x²) is​

Answer 1

$\textbf{Given:}$

$\mathsf{sin(x^2)}$

$\textbf{To find:}$

$\textsf{Coefficient of}\mathsf{x^8\;in\;the\;expansion\;of\;sin(x^2)}$

$\textbf{Solution:}$

$\mathsf{We\;know\;that,\;the\;series\;expansion\;of\;sin\,x\;is}$

$\boxed{\mathsf{sinx=\dfrac{x^1}{1!}-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\;.\;.\;.\;.\;.\;.}}$

$\mathsf{Replace\;x\;by\;x^2,\;we\;get}$

$\mathsf{sin(x^2)=\dfrac{(x^2)^1}{1!}-\dfrac{(x^2)^3}{3!}+\dfrac{(x^2)^5}{5!}-\;.\;.\;.\;.\;.\;.}$

$\mathsf{sin(x^2)=\dfrac{x^2}{1!}-\dfrac{x^6}{3!}+\dfrac{x^{10}}{5!}-\;.\;.\;.\;.\;.\;.}$

$\mathsf{Clearly\;there\;is\;no\;term\;contining\;x^8\;in\;the\;expansion\;of\;sin(x^2)}$

$\underline{\mathsf{Hence,\;Coefficient\;of\;X^8\;is\;0}}$

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