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Question Number 119773 by bemath last updated on 26/Oct/20

∫_(−4) ^4  x^3 (√(16−x^2  )) sec x dx

$$\underset{−\mathrm{4}} {\overset{\mathrm{4}} {\int}}\:{x}^{\mathrm{3}} \sqrt{\mathrm{16}−{x}^{\mathrm{2}} \:}\:\mathrm{sec}\:{x}\:{dx}\: \\ $$

Answered by MJS_new last updated on 27/Oct/20

f(x)=x^3 (√(16−x^2 ))sec x is odd ⇒ answer is 0

$${f}\left({x}\right)={x}^{\mathrm{3}} \sqrt{\mathrm{16}−{x}^{\mathrm{2}} }\mathrm{sec}\:{x}\:\mathrm{is}\:\mathrm{odd}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{0} \\ $$

Answered by john santu last updated on 27/Oct/20

i think integral divergent

$${i}\:{think}\:{integral}\:{divergent} \\ $$

Answered by mathmax by abdo last updated on 27/Oct/20

if sec x=(1/(cosx)) ⇒∫_(−4) ^4 (x^3 /(cosx))(√(16−x^2 ))dx =0  becsuse the function under integral is odd

$$\mathrm{if}\:\mathrm{sec}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{cosx}}\:\Rightarrow\int_{−\mathrm{4}} ^{\mathrm{4}} \frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{cosx}}\sqrt{\mathrm{16}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=\mathrm{0} \\ $$$$\mathrm{becsuse}\:\mathrm{the}\:\mathrm{function}\:\mathrm{under}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{odd} \\ $$

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