Question

Integral of x cube of sin x power of 4 of dx

Answer 1

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$I = \frac{3x^4}{32}+ \frac{x^3}{32}[sin(4x) -8sin(2x)]$

$\:\:\:\:\: - \frac{3x^2}{128}[16cos(2x) - cos(4x)]$

$\:\:\:\:\:- \frac{6x}{128}[8sin(2x) - \frac{sin(4x)}{4}]$

$\:\:\:\:\:+ \frac{6}{128} [\frac{cos(4x)}{16} - 4cos(2x)]$

$\green{\underline{</p><h2><strong><u>★</u></strong><strong><u>EXPLAINATION</u></strong><strong><u>★</u></strong></h2><p>}}$

$I = \int x^3sin^4(x)dx$

Using integration by parts,

$I = x^3\int sin^4(x)dx$

$\:\:\:\:\:- \int[\frac{d}{dx}(x^3)\int sin^4(x)dx]dx$

$I = x^3I_1(x) - \int[3x^2I_1(x)]dx$

$I = x^3I_1(x) - 3\int[x^2I_1(x)]dx$

Now

$I_1 = \int sin^4(x)dx$

$I_1 = \frac{1}{4}\int [2sin^2(x)]^2 dx$

$I_1 = \frac{1}{4}\int [1 - cos(2x)]^2 dx$

$I_1 = \frac{1}{4}\int [1 - 2cos(2x) + cos^2(2x)]dx$

$I_1 = \frac{1}{4}\int [1 - 2cos(2x) + \frac{1 + cos(4x)}{2}]dx$

$I_1 = \frac{1}{4}\int [\frac{3}{2}- 2cos(2x) + \frac{cos(4x)}{2}]dx$

$I_1(x) = \frac{1}{4}[\frac{3x}{2} - \frac{2sin(2x)}{2} + \frac{sin(4x)}{8}]dx$

$I_1(x) =\frac{3x}{8} + \frac{1}{32}[sin(4x) -8sin(2x)]$

$I_1(x) =\frac{3x}{8} + f(sin(x))$

Putting value of $I_1(x),$ we get

$I = x^3I_1(x) - 3\int[x^2[\frac{3x}{8}$

$\:\:\:\:\:+ f(sin(x))] ]dx$

$I = x^3[\frac{3x}{8} + f(sin(x))]$

$\:\:\:\:\:- \frac{9}{8}\int x^3dx - 3\int [x^2 f(sin(x))]dx$

$I = \frac{3x^4}{8}+ x^3f(sin(x))$

$\:\:\:\:\:- \frac{9x^4}{32} - 3\int x^2 f(sin(x))]dx$

$I = \frac{3x^4}{32}+ x^3f(sin(x))$

$\:\:\:\:\:- 3\int [x^2 f(sin(x))]dx$

Using integration by parts in I again,

$I = \frac{3x^4}{32}+ x^3f(sin(x))$

$\:\:\:\:\:- 3[x^2\int f(sin(x))dx$

$\:\:\:\:\:- \int [\frac{d}{dx}(x^2)\int f(sin(x))dx] dx$

$I = \frac{3x^4}{32}+ x^3f(sin(x))$

$\:\:\:\:\:- 3x^2\int f(sin(x))dx$

$\:\:\:\: +6\int [x\int f(sin(x))dx]dx$

$I = \frac{3x^4}{32}+ x^3f(sin(x))$

$\:\:\:\:\:- 3x^2I_2(x) + 6\int [xI_2(x)]dx$

Now,

$I_2(x) = \int f(sin(x))dx$

$I_2(x) = \int \frac{1}{32}[sin(4x) -8sin(2x)]dx$

$I_2(x) = \frac{1}{32}\int[sin(4x) -8sin(2x)]dx$

$I_2(x) = \frac{1}{32}[\frac{-cos(4x)}{4} -8\frac{-cos(2x)}{2}]dx$

$I_2(x) = \frac{1}{32}[-\frac{cos(4x)}{4} +4cos(2x)]dx$

$I_2(x) = \frac{1}{128}[16cos(2x) - cos(4x)]$

Now

Apply integration by part again in I,

$I = \frac{3x^4}{32}+ x^3f(sin(x))$

$\:\:\:\:\:- 3x^2I_2(x) + 6x\int I_2(x)dx$

$\:\:\:\:\:- 6\int[\frac{d}{dx}(x)\int I_2(x)dx]dx$

$I = \frac{3x^4}{32}+ x^3f(sin(x))$

$\:\:\:\:\:- 3x^2I_2(x) + 6xI_3(x)$

$\:\:\:\:\:- 6\int[I_3(x)]dx$

$I = \frac{3x^4}{32}+ x^3f(sin(x))$

$\:\:\:\:- 3x^2I_2(x) + 6xI_3(x) - 6I_4(x)$

where

$I_3(x) = \int I_2(x)dx$

$I_4(x) = \int I_3(x)dx$

Now

$I_3(x) = \int I_2(x)dx$

$I_3(x) = \frac{1}{128}\int [16cos(2x)$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: - cos(4x)]dx$

$I_3(x) = \frac{1}{128}[16\frac{sin(2x)}{2} - \frac{sin(4x)}{4}]$

$I_3(x) = \frac{1}{128}[8sin(2x) - \frac{sin(4x)}{4}]$

And

$I_4(x) = \int I_3(x)dx$

$I_4(x) = \frac{1}{128}\int [8sin(2x)- \frac{sin(4x)}{4}]dx$

$I_4(x) = \frac{1}{128} [8\frac{-cos(2x)}{2}- \frac{-cos(4x)}{16}]$

$I_4(x) = \frac{1}{128} [\frac{cos(4x)}{16} - 4cos(2x)]$

Hence

$I = \frac{3x^4}{32}+ \frac{x^3}{32}[sin(4x) -8sin(2x)]$

$\:\:\:\:\:- \frac{3x^2}{128}[16cos(2x) - cos(4x)]$

$\:\:\:\:\:- \frac{6x}{128}[8sin(2x) - \frac{sin(4x)}{4}]$

$\:\:\:\:\:+ \frac{6}{128} [\frac{cos(4x)}{16} - 4cos(2x)]$