Question-185058

Question Number 185058 by saboorhalimi last updated on 16/Jan/23

Answered by Frix last updated on 16/Jan/23

kcos nx +msin nx =(√(k^2 +m^2 ))sin (nx+tan^(−1) (k/m)) Ω=(√(k^2 +m^2 ))∫_0 ^π ∣sin (nx+tan^(−1) (k/m))∣dx= =((√(k^2 +m^2 ))/n)×∫_(tan^(−1) (k/m)) ^(nπ+tan^(−1) (k/m)) ∣sin t∣ dt= =((√(k^2 +m^2 ))/n) ×∫_0 ^(nπ) ∣sin t∣dt=(√(k^2 +m^2 ))∫_0 ^π sin t dt= =2(√(k^2 +m^2 ))

$${k}\mathrm{cos}\:{nx}\:+{m}\mathrm{sin}\:{nx}\:=\sqrt{{k}^{\mathrm{2}} +{m}^{\mathrm{2}} }\mathrm{sin}\:\left({nx}+\mathrm{tan}^{−\mathrm{1}} \:\frac{{k}}{{m}}\right) \\ $$$$\Omega=\sqrt{{k}^{\mathrm{2}} +{m}^{\mathrm{2}} }\underset{\mathrm{0}} {\overset{\pi} {\int}}\mid\mathrm{sin}\:\left({nx}+\mathrm{tan}^{−\mathrm{1}} \:\frac{{k}}{{m}}\right)\mid{dx}= \\ $$$$=\frac{\sqrt{{k}^{\mathrm{2}} +{m}^{\mathrm{2}} }}{{n}}×\underset{\mathrm{tan}^{−\mathrm{1}} \:\frac{{k}}{{m}}} {\overset{{n}\pi+\mathrm{tan}^{−\mathrm{1}} \:\frac{{k}}{{m}}} {\int}}\mid\mathrm{sin}\:{t}\mid\:{dt}= \\ $$$$=\frac{\sqrt{{k}^{\mathrm{2}} +{m}^{\mathrm{2}} }}{{n}}\:×\underset{\mathrm{0}} {\overset{{n}\pi} {\int}}\mid\mathrm{sin}\:{t}\mid{dt}=\sqrt{{k}^{\mathrm{2}} +{m}^{\mathrm{2}} }\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{sin}\:{t}\:{dt}= \\ $$$$=\mathrm{2}\sqrt{{k}^{\mathrm{2}} +{m}^{\mathrm{2}} } \\ $$

Commented by saboorhalimi last updated on 16/Jan/23

$$\:\:\:\:\boldsymbol{{nice}}\:\boldsymbol{{solution}} \\ $$

Commented by Frix last updated on 16/Jan/23

$$\mathrm{Thank}\:\mathrm{you},\:\mathrm{I}\:\mathrm{added}\:\mathrm{some}\:\mathrm{lines}. \\ $$

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Question-185058

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