cos-4-x-cos-4-2x-dx-

Question Number 118436 by bramlexs22 last updated on 17/Oct/20

\int \cos^{4} (x) \cos^{4} (2 x) d x

Answered by benjo_mathlover last updated on 17/Oct/20

(1) \cos^{4} (x) = {(\frac{1}{2} + \frac{1}{2} \cos (2 x))}^{2}

= \frac{1}{4} + \frac{1}{2} \cos (2 x) + \frac{1}{4} (\frac{1}{2} + \frac{1}{2} \cos (4 x))

= \frac{3}{8} + \frac{1}{2} \cos (2 x) + \frac{1}{8} \cos (4 x)

(2) \cos^{4} (2 x) = {(\frac{1}{2} + \frac{1}{2} \cos (4 x))}^{2}

= \frac{1}{4} + \frac{1}{2} \cos (4 x) + \frac{1}{4} (\frac{1}{2} + \frac{1}{2} \cos (8 x))

= \frac{3}{8} + \frac{1}{2} \cos (4 x) + \frac{1}{8} \cos (8 x)

T h e n (1) \times (2) :

= \frac{9}{64} + \frac{3}{16} \cos (4 x) + \frac{3}{64} \cos (8 x) + \frac{3}{16} \cos (2 x)

+ \frac{1}{8} \cos (6 x) + \frac{1}{8} \cos (2 x) + \frac{1}{32} \cos (10 x)

+ \frac{1}{32} \cos (6 x) + \frac{3}{64} \cos (4 x) + \frac{1}{32} + \frac{1}{32} \cos (8 x)

+ \frac{1}{128} \cos (12 x) + \frac{1}{128} \cos (4 x)

Answered by mathmax by abdo last updated on 17/Oct/20

I = \int \cos^{4} x \cos^{4} (2 x) dx \Rightarrow I = \int {(\frac{1 + \cos (2 x)}{2})}^{2} {(\frac{1 + \cos (4 x)}{2})}^{2} dx

= \frac{1}{16} \int {(1 + \cos (2 x))}^{2} {(1 + \cos (4 x))}^{2} dx

= \frac{1}{16} \int (1 + 2 \cos (2 x) + \cos^{2} (2 x)) (1 + 2 \cos (4 x) + \cos^{2} (4 x)) dx

16 I = \int (1 + 2 \cos (4 x) + \cos^{2} (4 x) + 2 \cos (2 x) + 4 \cos (2 x) \cos (4 x)

+ 2 \cos (2 x) \cos^{2} (4 x) + \cos^{2} (2 x) + {2 \cos}^{2} (2 x) \cos (4 x) + \cos^{2} (2 x) \cos^{2} (4 x)) dx

= x + \frac{1}{2} \sin (4 x) + \int \cos^{2} (4 x) dx + \sin (2 x) + 4 \int \cos (2 x) \cos (4 x) dx

+ 2 \int \cos (2 x) \cos^{2} (4 x) dx + \int \cos^{2} (2 x) dx + 2 \int \cos^{2} (2 x) \cos (4 x) dx

+ \int \cos^{2} (2 x) \cos^{2} (4 x) dx we have

\int \cos^{2} (4 x) dx = \int \frac{1 + \cos (8 x)}{2} dx = \frac{x}{2} + \frac{1}{16} \sin (8 x) + c_{0}

\int \cos (2 x) \cos (4 x) dx = \frac{1}{2} \int (\cos (6 x) + \cos (2 x)) dx

= \frac{1}{12} \sin (6 x) + \frac{1}{4} \sin (2 x) + c_{1}

\int \cos (2 x) \cos^{2} (4 x) dx = \frac{1}{2} \int \cos (2 x) (1 + \cos (8 x)) dx

= \frac{1}{4} \sin (2 x) + \frac{1}{2} \int \cos (2 x) \cos (8 x) dx

= \frac{1}{4} \sin (2 x) + \frac{1}{4} \int (\cos (10 x) + \cos (6 x)) dx

= \frac{1}{4} \sin (2 x) + \frac{1}{40} \sin (10 x) + \frac{1}{24} \sin (6 x) + c_{2}

\int \cos^{2} (2 x) dx = \frac{1}{2} \int (1 + \cos (4 x)) dx = \frac{x}{2} + \frac{1}{8} \sin (4 x) + c_{3}

\int \cos^{2} (2 x) \cos (4 x) dx = \frac{1}{2} \int (1 + \cos (4 x)) \cos 4 xdx

= \frac{1}{2} \int \cos (4 x) dx + \frac{1}{4} \int (1 + \cos (8 x)) dx

= \frac{1}{8} \sin (4 x) + \frac{x}{4} + \frac{1}{32} \sin (8 x) + c_{4} also

\int \cos^{2} (2 x) \cos^{2} (4 x) dx = \frac{1}{4} \int (1 + \cos (4 x)) (1 + \cos (8 x)) dx

= \frac{1}{4} \int (1 + \cos (8 x) + \cos (4 x) + \cos (4 x) \cos (8 x)) dx

= \frac{x}{4} + \frac{1}{32} \sin (8 x) + \frac{1}{16} \sin (4 x) + \frac{1}{8} \int \cos (12 x) + \cos (4 x)) dx

= \frac{x}{4} + \frac{1}{32} \sin (8 x) + \frac{1}{16} \sin (4 x) + \frac{1}{8.12} \sin (12 x) + \frac{1}{32} \sin (4 x)

rest to collect the integrals . .

Answered by MJS_new last updated on 17/Oct/20

\int \cos^{4} x \cos^{4} 2 x d x =

[t = \tan x \to d x = \cos^{2} x d t]

= \int \frac{{(t - 1)}^{4} {(t + 1)}^{4}}{{(t^{2} + 1)}^{7}} d t =

[{Ostrogradski}^{'} s Method]

= \frac{t (165 t^{10} + 935 t^{8} + 1986 t^{6} + 3006 t^{4} + 1305 t^{2} + 795)}{960 {(t^{2} + 1)}^{6}} +

+ \frac{11}{64} \int \frac{d t}{t^{2} + 1} [which = \frac{11}{64} \arctan t]

now put t = \tan x

Answered by 1549442205PVT last updated on 17/Oct/20

\cos^{4} x = {(\frac{1 + \cos 2 x}{2})}^{2} = \frac{1 + 2 \cos 2 x + \cos^{2} 2 x}{4}

Put F = \int \cos^{4} (x) \cos^{4} (2 x) dx \Rightarrow 4 F =

\int \cos^{4} 2 xdx + 2 \int \cos^{4} 2 xcos 2 xdx + \int \cos^{6} 2 xdx

\cos^{4} 2 x = {(\frac{1 + \cos 4 x}{2})}^{2} = \frac{1 + 2 \cos 4 x + \cos^{2} 4 x}{4}

= \frac{1}{4} (1 + 2 \cos 4 x + \frac{1 + \cos 8 x}{2}) (1)

= {(1 - \sin^{2} 2 x)}^{2} = 1 - {2 \sin}^{2} 2 x + \sin^{4} 2 x (2)

{\cos^{6} 2 x = \frac{1 + \cos 4 x}{2})}^{3} = \frac{1}{8} (1 + 3 \cos 4 x

+ {3 \cos}^{2} 4 x + \cos^{3} 4 x)

= \frac{1}{8} + \frac{3}{8} \cos 4 x + \frac{3}{16} (1 + \cos 8 x) + (1 - \sin^{2} 4 x) \cos 4 x

= \frac{5}{16} + \frac{3}{8} \cos 4 x + \frac{3}{16} \cos 8 x + (1 - {2 \sin}^{2} 4 x + \sin^{4} 4 x) \cos 4 x (3)

From (1) (2) (3) we get

4 F = \int \frac{11}{16} dx + \frac{7}{8} \int \cos 4 xdx + \frac{5}{16} \int \cos 8 xdx

+ \frac{1}{4} \int (1 - \sin^{2} 4 x) d (\sin 4 x)

+ \int (1 - {2 \sin}^{2} 2 x + \sin^{4} 2 x) d (\sin 2 x)

= \frac{11}{16} x + \frac{7}{32} \sin 4 x + \frac{5}{128} \sin 8 x + \frac{1}{4} \sin 4 x

- \frac{1}{12} \sin^{3} 4 x + \sin 2 x - \frac{2}{3} \sin^{3} 2 x + \frac{1}{5} \sin^{5} 2 x

F = \frac{1}{4} (\frac{11}{16} x + \frac{7}{32} \sin 4 x + \frac{5}{128} \sin 8 x

- \frac{1}{12} \sin^{3} 4 x + \sin 2 x - \frac{2}{3} \sin^{3} 2 x

+ \frac{1}{5} \sin^{5} 2 x) + C