question_answer

Find the value of \[\int_{-\pi /4}^{\pi /4}{({{x}^{3}}+{{x}^{4}}+{{\tan }^{3}}x)\,dx.}\] 

Answer:

Let \[l=\int_{-\pi /4}^{\pi /4}{({{x}^{3}}+{{x}^{4}}+{{\tan }^{3}}x)dx}\] and \[f(x)={{x}^{3}}+{{\tan }^{3}}x,\] then \[f(-\,x)={{(-\,x)}^{3}}+{{\tan }^{3}}(-x)=-\,{{x}^{3}}-\,{{\tan }^{3}}x=-f(x)\] \[\therefore \] f(x) is an odd function.                      We know that if \[\phi (x)\] is an odd function,              then \[\int_{-\,a}^{a}{\phi (x)\,dx=0}\] and \[\phi (x)\] is even function, then \[\int_{-\,a}^{a}{\phi (x)\,dx=2\int_{0}^{a}{\phi (x)dx}}\] \[\therefore \]      \[l=\int_{-\pi /4}^{\pi /4}{({{x}^{3}}+{{\tan }^{3}}x)\,dx}+\int_{-\pi /4}^{\pi /4}{{{x}^{4}}dx}\]             \[=0+2\int_{0}^{\pi /4}{{{x}^{4}}dx}\]             \[=2\left[ \frac{{{x}^{5}}}{5} \right]_{0}^{\pi /4}=\frac{2}{5}{{\left( \frac{\pi }{4} \right)}^{5}}=\frac{2}{5}\times \frac{{{\pi }^{5}}}{{{4}^{5}}}=\frac{{{\pi }^{5}}}{2560}\]