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Question: I![](https://cdn.pureessence.tech/canvas_252.png?top_left_x=762&top_left_y=0&width=200&height=300)(c...

I(cos x)4 dx = Ax + B sin 2x + C sin4x then {A, B, C}

equals-

A

{38,14,132}\left\{ \frac{3}{8},\frac{1}{4},\frac{1}{32} \right\}

B

{38,12,16}\left\{ \frac{3}{8},\frac{1}{2},\frac{1}{6} \right\}

C

{34,14,116}\left\{ \frac{3}{4},\frac{1}{4},\frac{1}{16} \right\}

D

{38,14,132}\left\{ \frac{3}{8},\frac{1}{4},\frac{1}{32} \right\}

Answer

{38,14,132}\left\{ \frac{3}{8},\frac{1}{4},\frac{1}{32} \right\}

Explanation

Solution

I = ∫(1+cos⁡2x2)2\int_{}^{}\left( \frac{1 + \cos 2x}{2} \right)^{2}dx

= 14\frac{1}{4} ∫(1+cos⁡22x+2cos⁡2x)\int_{}^{}{(1 + \cos^{2}2x + 2\cos 2x)}dx

= 14\frac { 1 } { 4 } ∫(1+1+cos⁡4x2+2cos⁡2x)\int_{}^{}\left( 1 + \frac{1 + \cos 4x}{2} + 2\cos 2x \right)dx

= 18\frac { 1 } { 8 } ∫(3+cos⁡4x+4cos⁡2x)dx\int_{}^{}{(3 + \cos 4x + 4\cos 2x)dx}

= 3x8\frac { 3 x } { 8 }+ sin⁡4x32\frac{\sin 4x}{32}+ 4sin⁡2x8×2\frac{4\sin 2x}{8 \times 2}