Question

$Evaluate:{\text{ }}\int {{{\sin }^4}x{{\cos }^4}x} dx$

Answer 1

Hint : In this integration question, the trigonometry identities must be known to solve this problem. The key observation here to use the formula ${\cos ^2}x = \dfrac{{1 + \cos 2x}}{2}$ and also ${\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}$ . By using these two identities the integrand can be simplified and the integration can be done easily.

Complete step-by-step answer :
$Let{\text{ I = }}\int {{{\sin }^4}x{{\cos }^4}x} dx$
Multiplying and dividing by 16,
${\text{I = }}\int {\left( {\dfrac{{16}}{{16}}} \right){{\sin }^4}x{{\cos }^4}x} dx$
$\Rightarrow {\text{I = }}\dfrac{1}{{16}}\int {16{{\sin }^4}x{{\cos }^4}x} dx$
$\because \sin 2x = 2\sin x\cos x$
$\Rightarrow {\text{I = }}\dfrac{1}{{16}}\int {{{\left( {\sin 2x} \right)}^4}} dx$
$\because {\sin ^2}x = 1 - {\cos ^2}x$
$\Rightarrow {\text{I = }}\dfrac{1}{{16}}\int {{{\left( {1 - {{\cos }^2}2x} \right)}^2}} dx$
On expanding,
$\Rightarrow {\text{I = }}\dfrac{1}{{16}}\int {\left( {1 + {{\cos }^4}2x - 2{{\cos }^2}2x} \right)} dx$
$\because {\cos ^2}x = \dfrac{{1 + \cos 2x}}{2}$
Or, ${\cos ^4}x = {\left( {\dfrac{{1 + \cos 2x}}{2}} \right)^2}$
$\therefore {\text{I = }}\dfrac{1}{{16}}\int {\left( {1 + {{\left( {\dfrac{{1 + \cos 4x}}{2}} \right)}^2} - 2\left( {\dfrac{{1 + \cos 4x}}{2}} \right)} \right)} dx$
On expanding and simplifying further,
$\therefore {\text{I = }}\dfrac{1}{{16}}\int {\left( {1 + \left( {\dfrac{{1 + {{\cos }^2}4x + 2\cos 4x}}{4}} \right) - 1 - \cos 4x} \right)} dx$
On taking LCM,
$\therefore {\text{I = }}\dfrac{1}{{16}}\int {\dfrac{{\left( {4 + 1 + {{\cos }^2}2x + 2\cos 2x - 4 - 4\cos 2x} \right)}}{4}} dx$
On simplifying further,
$\therefore {\text{I = }}\dfrac{1}{{16}}\int {\dfrac{{\left( {1 + {{\cos }^2}4x - 2\cos 4x} \right)}}{4}} dx$
Or,
$\therefore {\text{I = }}\dfrac{1}{{64}}\int {\left( {1 + {{\cos }^2}4x - 2\cos 4x} \right)} dx$
Again using ${\cos ^2}x = \dfrac{{1 + \cos 2x}}{2}$
${\text{I = }}\dfrac{1}{{64}}\int {\left( {1 + \dfrac{{1 + \cos 8x}}{2} - 2\cos 4x} \right)} dx$
On taking LCM,
$\therefore {\text{I = }}\dfrac{1}{{64}}\int {\dfrac{{\left( {2 + 1 + \cos 8x - 4\cos 4x} \right)}}{2}} dx$
On simplifying further,
${\text{I = }}\dfrac{1}{{128}}\int {\left( {3 + \cos 8x - 4\cos 4x} \right)} dx$
On separating the terms,
${\text{I = }}\dfrac{1}{{128}}\left( {\int {3dx + \int {\cos 8xdx - 4\int {\cos 4x} } } } \right)$
$\Rightarrow {\text{I = }}\dfrac{1}{{128}}\left( {3x + {c_1} + \left( { - \dfrac{{\sin 8x}}{8} + {c_2}} \right) - 4\left( { - \dfrac{{\sin 4x}}{4} + {c_3}} \right)} \right)$
Where ${c_1},{\text{ }}{c_2}{\text{ }}and{\text{ }}{c_3}$ are the constants of integration.
After simplifying and rearranging the terms,
$\Rightarrow {\text{I = }}\dfrac{1}{{128}}\left( {3x + \left( { - \dfrac{{\sin 8x}}{8}} \right) + \left( {\sin 4x} \right) + C} \right)$
Where $C = {c_1}{\text{ + }}{c_2} - 4{c_3}$

Note : The value of the constant can be found if the initial condition is given i.e. when x=0, the result of the integration would be some constant value, and by equating the value of the constant of integration can be found. This is indefinite-integral hence the constant of integration must be there but if it is definite-integral then the final answer would be some constant value. Also the results of the standard indefinite integral must be known so that it can be used directly.