Question

How do you find the integral of ${{\sin }^{3}}\left[ x \right]dx$?

Answer 1

In this problem, we have to find the integral of ${{\sin }^{3}}\left[ x \right]dx$. We can now split the given integral into two parts. We can integrate the first integral as it will have only one function. We can then integrate the second integral using the substitution method. We can then combine the value of the first and the second integral value to get the final integral value.

Complete step-by-step answer:
Here we have to find the integral of ${{\sin }^{3}}\left[ x \right]dx$.
We can now write the given integral in the form,
$\Rightarrow \int{\sin x\left( {{\sin }^{2}}x \right)}dx$
We know that ${{\sin }^{2}}x=1-{{\cos }^{2}}x$, we can now substitute this in the above step and multiply, we get

& \Rightarrow \int{\sin x\left( 1-{{\cos }^{2}}x \right)dx} \\\ & \Rightarrow \int{\sin x-\sin x{{\cos }^{2}}x} \\\ \end{aligned}$$ We can now split the integral into two parts, we get $$\Rightarrow \int{\sin xdx-\int{\sin x{{\cos }^{2}}xdx}}$$ ……. (1) We can now integrate the first integral, we get $$\Rightarrow \int{\sin xdx}=-\cos x$$……. (2) We can now integrate the second integral, using substitution, Let, $$\begin{aligned} & u=\cos x \\\ & du=-\sin xdx \\\ \end{aligned}$$ We can now substitute the above step in second integral, we get $$\Rightarrow \int{{{u}^{2}}du}$$ We can now integrate the above step, we get $$\Rightarrow \dfrac{{{u}^{3}}}{3}+C$$ We can now substitute the value of u in the above step, $$\Rightarrow \dfrac{1}{3}{{\cos }^{3}}x+C$$…….. (3) We can now substitute (2) and (3) in (1), we get $$\Rightarrow \dfrac{1}{3}{{\cos }^{3}}x-\cos x+C$$ Therefore, the integral of $${{\sin }^{3}}\left[ x \right]dx=\dfrac{1}{3}{{\cos }^{3}}x-\cos x+C$$. **Note:** We should always remember the integral formulas, such as integral of $$\sin x=-\cos x$$ and differentiation of $$\cos x=-\sin x$$ . We should know that after using the substitution method and integrating the problem, we should have to substitute the original value we have assigned before.