The integration of $\sin^3{x}$ function with respect to $x$ can be calculated in integral calculus in two methods.

The sin and cos functions both have direct relation in integral calculus. If the $sin^3{x}$ function is expressed in terms of both sine and cosine functions. The integral of $\sin^3{x}$ with respect to $x$ can be done easily in mathematics.

The sin cubed function can be written in terms of cos and sin functions by splitting them as two factors.

$= \,\,\,$ $\displaystyle \int{\sin^2{x} \times \sin{x}}dx$

Actually, the square of sin function can be written in terms of cos function by the sin squared formula.

$= \,\,\,$ $\displaystyle \int{(1-\cos^2{x}) \times \sin{x}}dx$

$= \,\,\,$ $\displaystyle \int{(1-\cos^2{x})\sin{x}}dx$

The differentiation of cos function is equal to minus of sin function. Hence, take $t = \cos{x}$, then $dt = -\sin{x}dx$. Therefore, $\sin{x}dx = -dt$.

Now, eliminate the $x$ terms by $t$ terms.

$= \,\,\,$ $\displaystyle \int{(1-t^2})(-dt)$

$= \,\,\,$ $-\displaystyle \int{(1-t^2})dt$

$= \,\,\,$ $\displaystyle \int{(-1+t^2})dt$

There is no straight integral rule to find the integration of difference of the two terms. So, use difference rule of integration.

$= \,\,\,$ $\displaystyle \int{(-1)}dt+\displaystyle \int{t^2}dt$

$= \,\,\,$ $-\displaystyle \int dt+\displaystyle \int{t^2}dt$

The integration of both functions can be done by using integral of one rule and power rule of integration.

$= \,\,\,$ $-t+\dfrac{t^3}{3}+C$

The actual function is not in terms of $t$. So, replace back the $t$ by its assumed value.

$= \,\,\,$ $-\cos{x}+\dfrac{\cos^3{x}}{3}+C$

$= \,\,\,$ $\dfrac{\cos^3{x}}{3}-\cos{x}+C$

The integration of $\sin^3{x}$ with respect to $x$ can be done by transforming the sin cubed function as per sin triple angle formula.

Actually, there is no formula for finding the integral of sin cubed function in integral calculus. So, an alternative approach must be used to start finding the sine cubed function. In this method, sin triple angle formula is used to express sin cubed function in terms of sine and sin triple angle functions.

$\sin{3x} \,=\, 3\sin{x}-4\sin^3{x}$

$\implies 4\sin^3{x} \,=\, 3\sin{x}-\sin{3x}$

$\implies \sin^3{x} \,=\, \dfrac{3\sin{x}-\sin{3x}}{4}$

$= \,\,\,$ $\displaystyle \int{\Bigg[\dfrac{3\sin{x}-\sin{3x}}{4}\Bigg]}dx$

$= \,\,\,$ $\displaystyle \int{\Bigg[\dfrac{3}{4}\sin{x}-\dfrac{1}{4}\sin{3x}\Bigg]}dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{3}{4}\sin{x}}dx-\displaystyle \int{\dfrac{1}{4}\sin{3x}}dx$

$= \,\,\,$ $\displaystyle \dfrac{3}{4}\int{\sin{x}}dx-\displaystyle \dfrac{1}{4}\int{\sin{3x}}dx$

The integration of first term can be calculated as per integral formula of sin function but it is not possible to find the integration of $\sin{3x}$ function directly by this formula. However, it can be done by transforming the angle $3x$ in terms of a variable.

If $t = 3x$, then $dt = 3dx$. Therefore, $dx = \dfrac{dt}{3}$

$= \,\,\,$ $\displaystyle \dfrac{3}{4}\int{\sin{x}}dx-\displaystyle \dfrac{1}{4}\int{\sin{t}}\dfrac{dt}{3}$

$= \,\,\,$ $\displaystyle \dfrac{3}{4}\int{\sin{x}}dx-\displaystyle \dfrac{1}{4 \times 3}\int{\sin{t}}dt$

$= \,\,\,$ $\displaystyle \dfrac{3}{4}\int{\sin{x}}dx-\displaystyle \dfrac{1}{12}\int{\sin{t}}dt$

There are two integral terms in which one term is in terms of $x$ and other term is in terms of $t$ but both terms represents the indefinite integral of sin function.

$= \,\,\,$ $\dfrac{3}{4}(-\cos{x})-\dfrac{1}{12}(-\cos{t})+C$

$= \,\,\,$ $-\dfrac{3}{4}\cos{x}+\dfrac{1}{12}\cos{t}+C$

The second term in this expression is in terms of $t$ but our actual function is in terms of $x$. So, replace the value of $t$ in terms $x$.

$= \,\,\,$ $-\dfrac{3}{4}\cos{x}+\dfrac{1}{12}\cos{3x}+C$

The first term is cos function but second term is cos triple angle function in the expression. It can be simplified further by expanding $\cos{3x}$ function in terms of $\cos{x}$ by cos triple angle formula.

$= \,\,\,$ $-\dfrac{3}{4}\cos{x}+\dfrac{1}{12}\Bigg[4\cos^3{x}-3\cos{x}\Bigg]+C$

$= \,\,\,$ $-\dfrac{3}{4}\cos{x}+\dfrac{4}{12}\cos^3{x}-\dfrac{3}{12}\cos{x}+C$

$= \,\,\,$ $-\dfrac{3}{4}\cos{x}+\require{cancel} \dfrac{\cancel{4}}{\cancel{12}}\cos^3{x}-\require{cancel} \dfrac{\cancel{3}}{\cancel{12}}\cos{x}+C$

$= \,\,\,$ $-\dfrac{3}{4}\cos{x}+\dfrac{1}{3}\cos^3{x}-\dfrac{1}{4}\cos{x}+C$

$= \,\,\,$ $-\dfrac{3}{4}\cos{x}-\dfrac{1}{4}\cos{x}+\dfrac{1}{3}\cos^3{x}+C$

$= \,\,\,$ $\Bigg[-\dfrac{3}{4}-\dfrac{1}{4}\Bigg]\cos{x}+\dfrac{1}{3}\cos^3{x}+C$

$= \,\,\,$ $-\dfrac{4}{4}\cos{x}+\dfrac{1}{3}\cos^3{x}+C$

$= \,\,\,$ $\require{cancel} -\dfrac{\cancel{4}}{\cancel{4}}\cos{x}+\dfrac{1}{3}\cos^3{x}+C$

$= \,\,\,$ $-\cos{x}+\dfrac{\cos^3{x}}{3}+C$

$= \,\,\,$ $\dfrac{\cos^3{x}}{3}-\cos{x}+C$

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