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

	Answered by tanmay last updated on 15/May/19
	$1>sin(3x)>−1 (x−1)≥x+sin(3x)≥x−1 ∫_0 ^π sin^4 (x+1)dx≥∫_0 ^π sin^4 (x+sin(3x))dx≥∫_0 ^π sin^4 (x−1)dx now I_1 =∫_0 ^π sin^4 (x+1)dx a=x+1 ∫_1 ^(π+1) sin^4 a da I_2 =∫_0 ^π sin^4 (x−1)dx b=x−1 ∫_(−1) ^(π−1) sin^4 bdb sin^4 θ=(((1−cos2θ)/2))^2 =((1−2cos2θ+((1+cos4θ)/2))/4) =((2−4cos2θ+1+cos4θ)/8) =((cos4θ−4cos2θ+3)/8) I_1 ≥I≥I_2 ∫_1 ^(π+1) sin^4 ada ∫_1 ^(π+1) ((cos4a−4cos2a+3)/8) =(1/8)×∣((sin4a)/4)−4×((sin2a)/2)+3a∣_1 ^(π+1) =(1/(32)){sin(4π+4)−sin4}−(1/4)×{sin(2π+2)−sin2}+(3/8)(π) =((3π)/8)←value of I_1 ∫_1 ^(π−1) sin^4 bdb (1/8)×∣((sin4b)/4)−4×((sin2b)/2)+3b∣_(−1) ^(π−1) =(1/(32)){sin(4π−4)−sin(−4)}−(1/4)×{sin(2π−2)−sin(−2)}+(3/8)(π−1−1) =(3/8)(π−2) so ((3π)/8)≥∫_0 ^π sin^4 (x+sin3x)dx≥(3/8)(π−2)$
	$1 > s i n (3 x) > - 1$ $(x - 1) ⩾ x + s i n (3 x) ⩾ x - 1$ $\int_{0}^{π} {s i n}^{4} (x + 1) d x ⩾ \int_{0}^{π} {s i n}^{4} (x + s i n (3 x)) d x ⩾ \int_{0}^{π} {s i n}^{4} (x - 1) d x$ $n o w$ $I_{1} = \int_{0}^{π} {s i n}^{4} (x + 1) d x$ $a = x + 1$ $\int_{1}^{π + 1} {s i n}^{4} a d a$ $I_{2} = \int_{0}^{π} {s i n}^{4} (x - 1) d x$ $b = x - 1$ $\int_{- 1}^{π - 1} {s i n}^{4} b d b$ ${s i n}^{4} θ = {(\frac{1 - c o s 2 θ}{2})}^{2}$ $= \frac{1 - 2 c o s 2 θ + \frac{1 + c o s 4 θ}{2}}{4}$ $= \frac{2 - 4 c o s 2 θ + 1 + c o s 4 θ}{8}$ $= \frac{c o s 4 θ - 4 c o s 2 θ + 3}{8}$ $I_{1} ⩾ I ⩾ I_{2}$ $\int_{1}^{π + 1} {s i n}^{4} a d a$ $\int_{1}^{π + 1} \frac{c o s 4 a - 4 c o s 2 a + 3}{8}$ $= \frac{1}{8} \times ∣ \frac{s i n 4 a}{4} - 4 \times \frac{s i n 2 a}{2} + 3 a ∣_{1}^{π + 1}$ $= \frac{1}{32} {s i n (4 π + 4) - s i n 4} - \frac{1}{4} \times {s i n (2 π + 2) - s i n 2} + \frac{3}{8} (π)$ $= \frac{3 π}{8} \leftarrow v a l u e o f I_{1}$ $\int_{1}^{π - 1} {s i n}^{4} b d b$ $\frac{1}{8} \times ∣ \frac{s i n 4 b}{4} - 4 \times \frac{s i n 2 b}{2} + 3 b ∣_{- 1}^{π - 1}$ $= \frac{1}{32} {s i n (4 π - 4) - s i n (- 4)} - \frac{1}{4} \times {s i n (2 π - 2) - s i n (- 2)} + \frac{3}{8} (π - 1 - 1)$ $= \frac{3}{8} (π - 2)$ $s o$ $\frac{3 π}{8} ⩾ \int_{0}^{π} {s i n}^{4} (x + s i n 3 x) d x ⩾ \frac{3}{8} (π - 2)$
	Commented by bhanukumarb2@gmail.com last updated on 16/May/19

	$b r i l l e n t m t h d t h a n k s s a l o t$
	Commented by tanmay last updated on 16/May/19

	$m o s t w e l c o m e . . .$
	Commented by bhanukumarb2@gmail.com last updated on 16/May/19

	$p l z s e e 59846 d o u b t$
	Commented by tanmay last updated on 16/May/19

	$o k . . . l e t m e t r y$
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