Question

Integrate the given function ${\sin ^4}x$

Answer 1

Hint: To integrate ${\sin ^4}x$ first we have to change the ${\sin ^4}x$ by using \[\cos 2\theta = 1 - 2{\sin ^2}\theta \] into ${\left( {\dfrac{{1 - \cos 2x}}{2}} \right)^2}$. It is also known that ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ so ${\left( {\dfrac{{1 - \cos 2x}}{2}} \right)^2} = \left( {\dfrac{{1 + {{\cos }^2}2x - 2\cos 2x}}{4}} \right)$. After that we will convert ${\cos ^2}2x$ by using \[\cos 2\theta = 2{\cos ^2}\theta - 1\] then on simplifying we get the value ${\sin ^4}x$ will be equal to $\dfrac{1}{8}\left( {3 + \cos 4x - 4\cos 2x} \right)$. Now we can integrate this. And according to the integration theorem we can separate the addition terms,$\int 1 dx = x$ and $\int {\cos } \theta = \sin \theta $.

Complete step-by-step answer:
$\int {\left( {{{\sin }^4}x} \right)} dx$
Converting ${\sin ^2}x$ by using \[\cos 2\theta = 1 - 2{\sin ^2}\theta \], we get
$ \Rightarrow \int {{{\left( {\dfrac{{1 - \cos 2x}}{2}} \right)}^2}} dx$
Now, factorize ${\left( {1 - \cos 2x} \right)^2}$ by using ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, we get
$ \Rightarrow \dfrac{1}{4}\int {\left( {1 + {{\cos }^2}2x - 2\cos 2x} \right)dx} $
Convert ${\cos ^2}2x$ by using \[\cos 2\theta = 2{\cos ^2}\theta - 1\], we get
$ \Rightarrow \dfrac{1}{4}\int {\left( {1 - 2\cos 2x + \dfrac{{\cos 4x + 1}}{2}} \right)} dx$
$ \Rightarrow \dfrac{1}{8}\int {\left( {3 + \cos 4x - 4\cos 2x} \right)} dx$
We can also write it as
$ \Rightarrow \dfrac{3}{8}\int {dx} + \dfrac{1}{8}\int {\left( {\cos 4x} \right)} dx - \dfrac{1}{2}\int {\left( {\cos 2x} \right)dx} $
It is known that $\int 1 dx = x$ and $\int {\cos } \theta = \sin \theta $, then on simplifying, we get
$ \Rightarrow \dfrac{{3x}}{8} + \dfrac{{\sin 4x}}{{32}} - \dfrac{{\sin 2x}}{4} + c$
$\therefore \int {\left( {{{\sin }^4}x} \right)} dx = \dfrac{{3x}}{8} + \dfrac{{\sin 4x}}{{32}} - \dfrac{{\sin 2x}}{4} + c$

Note: Trigonometric formula used in this integration are \[\cos 2\theta = 1 - 2{\sin ^2}\theta \],\[\cos 2\theta = 2{\cos ^2}\theta - 1\], whereas formula of integration uses are $\int 1 dx = x$ and $\int {\cos } \theta = \sin \theta $. Point to take care there is no any formula of direct integration of higher power of trigonometric ratio we have to change it into power 1 by using trigonometric formula. Constant in the integration can be taken out from the integration.