Question-59834

Question Number 59834 by bhanukumarb2@gmail.com last updated on 15/May/19

Answered by tanmay last updated on 15/May/19

1>sin(3x)>−1 (x−1)≥x+sin(3x)≥x−1 ∫_0 ^π sin^4 (x+1)dx≥∫_0 ^π sin^4 (x+sin(3x))dx≥∫_0 ^π sin^4 (x−1)dx now I_1 =∫_0 ^π sin^4 (x+1)dx a=x+1 ∫_1 ^(π+1) sin^4 a da I_2 =∫_0 ^π sin^4 (x−1)dx b=x−1 ∫_(−1) ^(π−1) sin^4 bdb sin^4 θ=(((1−cos2θ)/2))^2 =((1−2cos2θ+((1+cos4θ)/2))/4) =((2−4cos2θ+1+cos4θ)/8) =((cos4θ−4cos2θ+3)/8) I_1 ≥I≥I_2 ∫_1 ^(π+1) sin^4 ada ∫_1 ^(π+1) ((cos4a−4cos2a+3)/8) =(1/8)×∣((sin4a)/4)−4×((sin2a)/2)+3a∣_1 ^(π+1) =(1/(32)){sin(4π+4)−sin4}−(1/4)×{sin(2π+2)−sin2}+(3/8)(π) =((3π)/8)←value of I_1 ∫_1 ^(π−1) sin^4 bdb (1/8)×∣((sin4b)/4)−4×((sin2b)/2)+3b∣_(−1) ^(π−1) =(1/(32)){sin(4π−4)−sin(−4)}−(1/4)×{sin(2π−2)−sin(−2)}+(3/8)(π−1−1) =(3/8)(π−2) so ((3π)/8)≥∫_0 ^π sin^4 (x+sin3x)dx≥(3/8)(π−2)

$$\mathrm{1}>{sin}\left(\mathrm{3}{x}\right)>−\mathrm{1} \\ $$$$\:\left({x}−\mathrm{1}\right)\geqslant{x}+{sin}\left(\mathrm{3}{x}\right)\geqslant{x}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\pi} {sin}^{\mathrm{4}} \left({x}+\mathrm{1}\right){dx}\geqslant\int_{\mathrm{0}} ^{\pi} \:{sin}^{\mathrm{4}} \left({x}+{sin}\left(\mathrm{3}{x}\right)\right){dx}\geqslant\int_{\mathrm{0}} ^{\pi} {sin}^{\mathrm{4}} \left({x}−\mathrm{1}\right){dx} \\ $$$${now} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\pi} {sin}^{\mathrm{4}} \left({x}+\mathrm{1}\right){dx} \\ $$$${a}={x}+\mathrm{1} \\ $$$$\int_{\mathrm{1}} ^{\pi+\mathrm{1}} {sin}^{\mathrm{4}} {a}\:{da} \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\pi} {sin}^{\mathrm{4}} \left({x}−\mathrm{1}\right){dx} \\ $$$${b}={x}−\mathrm{1} \\ $$$$\int_{−\mathrm{1}} ^{\pi−\mathrm{1}} {sin}^{\mathrm{4}} {bdb} \\ $$$${sin}^{\mathrm{4}} \theta=\left(\frac{\mathrm{1}−{cos}\mathrm{2}\theta}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}−\mathrm{2}{cos}\mathrm{2}\theta+\frac{\mathrm{1}+{cos}\mathrm{4}\theta}{\mathrm{2}}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{2}−\mathrm{4}{cos}\mathrm{2}\theta+\mathrm{1}+{cos}\mathrm{4}\theta}{\mathrm{8}} \\ $$$$=\frac{{cos}\mathrm{4}\theta−\mathrm{4}{cos}\mathrm{2}\theta+\mathrm{3}}{\mathrm{8}} \\ $$$$\:{I}_{\mathrm{1}} \geqslant\boldsymbol{{I}}\geqslant{I}_{\mathrm{2}} \\ $$$$\int_{\mathrm{1}} ^{\pi+\mathrm{1}} {sin}^{\mathrm{4}} {ada} \\ $$$$\int_{\mathrm{1}} ^{\pi+\mathrm{1}} \frac{{cos}\mathrm{4}{a}−\mathrm{4}{cos}\mathrm{2}{a}+\mathrm{3}}{\mathrm{8}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}×\mid\frac{{sin}\mathrm{4}{a}}{\mathrm{4}}−\mathrm{4}×\frac{{sin}\mathrm{2}{a}}{\mathrm{2}}+\mathrm{3}{a}\mid_{\mathrm{1}} ^{\pi+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\left\{{sin}\left(\mathrm{4}\pi+\mathrm{4}\right)−{sin}\mathrm{4}\right\}−\frac{\mathrm{1}}{\mathrm{4}}×\left\{{sin}\left(\mathrm{2}\pi+\mathrm{2}\right)−{sin}\mathrm{2}\right\}+\frac{\mathrm{3}}{\mathrm{8}}\left(\pi\right) \\ $$$$=\frac{\mathrm{3}\pi}{\mathrm{8}}\leftarrow{value}\:{of}\:{I}_{\mathrm{1}} \\ $$$$\int_{\mathrm{1}} ^{\pi−\mathrm{1}} {sin}^{\mathrm{4}} {bdb} \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}×\mid\frac{{sin}\mathrm{4}{b}}{\mathrm{4}}−\mathrm{4}×\frac{{sin}\mathrm{2}{b}}{\mathrm{2}}+\mathrm{3}{b}\mid_{−\mathrm{1}} ^{\pi−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\left\{{sin}\left(\mathrm{4}\pi−\mathrm{4}\right)−{sin}\left(−\mathrm{4}\right)\right\}−\frac{\mathrm{1}}{\mathrm{4}}×\left\{{sin}\left(\mathrm{2}\pi−\mathrm{2}\right)−{sin}\left(−\mathrm{2}\right)\right\}+\frac{\mathrm{3}}{\mathrm{8}}\left(\pi−\mathrm{1}−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}\left(\pi−\mathrm{2}\right) \\ $$$$\boldsymbol{{so}}\: \\ $$$$\frac{\mathrm{3}\pi}{\mathrm{8}}\geqslant\int_{\mathrm{0}} ^{\pi} {sin}^{\mathrm{4}} \left({x}+{sin}\mathrm{3}{x}\right){dx}\geqslant\frac{\mathrm{3}}{\mathrm{8}}\left(\pi−\mathrm{2}\right) \\ $$$$ \\ $$

Commented by bhanukumarb2@gmail.com last updated on 16/May/19