## Calculus: Early Transcendentals 8th Edition

Need to prove $\int _{-1}^{1}(x^{5}-6x^{9}+\frac{sinx}{(1+x^{4})^{2}})dx= 0$ Consider $f(x)=x^{5}-6x^{9}+\frac{sinx}{(1+x^{4})^{2}}dx$ $f(-x)=(-x)^{5}-6(-x)^{9}+\frac{sin(-x)}{(1+(-x)^{4})^{2}}dx$ $f(-x)=-x^{5}+6x^{9}-\frac{sinx}{(1+x^{4})^{2}}dx=-f(x)$ Remember that : if $f(x)$ is an odd function that is $f(-x)$, then $\int _{-a}^{a}f(x) dx=0$ Therefore, $f(x)$ is odd and its integral from -1 to 1 is zero. Hence , the statement is true.