Question

Solve this integration: $\int{{{\sin }^{3}}x{{\cos }^{3}}xdx}$ using proper trigonometric identity and formula.

Answer 1

Hint: Multiply and divide the whole relation by 2 and use the trigonometric identity of $\sin 2x=2\sin x\cos x$ to get the given integral in simplified form. Use the trigonometric identity of $\sin 3x$, whenever required. It is given as $\sin 3x=3\sin x-4{{\sin }^{3}}x$ .

Complete step-by-step answer:

Let us suppose the value of the given integral be I. So, we get equation as
$I=\int{{{\sin }^{3}}x{{\cos }^{3}}xdx}$ …………………………(i)
Now, we can rewrite the equation (i) as
$I=\int{{{\left( \sin x\cos x \right)}^{3}}dx}$
Multiply and divide by 2 inside the bracket of RHS part of the above equation, we get
$I=\int{\dfrac{{{\left( 2\sin x\cos x \right)}^{3}}}{{{\left( 2 \right)}^{3}}}dx}$
$\Rightarrow I=\dfrac{1}{8}\int{{{\left( 2\sin x\cos x \right)}^{3}}dx}$ ………………………(ii)
Now, we know the trigonometric identity of $\sin 2x$ is given as
$\sin 2x=2\sin x\cos x$ ……………………………….(iii)
So, we can replace $2\sin x\cos x$ from the equation (iii) by $\sin 2x$ in the expression of equation (ii).
So, we get value of I as
$I=\dfrac{1}{8}\int{{{\left( \sin 2x \right)}^{3}}dx}$
$I=\dfrac{1}{8}\int{{{\sin }^{3}}2xdx}$ …...............................(iv)
As we know trigonometric identity of $\sin 3x$ can be given as
$\sin 3x=3\sin x-4{{\sin }^{3}}x$ ………………………..(v)
Now, put $x=2x$ to the equation (v), so, we get
$\sin 6x=3\sin 2x-4{{\sin }^{3}}2x$
Now, we can get value of ${{\sin }^{3}}2x$ as
$4{{\sin }^{3}}2x=3\sin 2x-\sin 6x$
$\Rightarrow {{\sin }^{3}}2x=\dfrac{3}{4}\sin 2x-\dfrac{1}{4}\sin 6x$ ………………………(vi)
Hence, using equation (vi), we can replace ${{\sin }^{3}}2x$ by $\dfrac{3}{4}\sin 2x-\dfrac{1}{4}\sin 6x$ . Hence, we get value of I as
$I=\dfrac{1}{8}\int{\left( \dfrac{3}{4}\sin 2x-\dfrac{1}{4}\sin 6x \right)dx}$
$I=\dfrac{3}{32}\int{\sin 2x-\dfrac{1}{32}\int{\sin 6x}dx}$
Now, we know $\int{\sin xdx=-\cos x}$
So, we get value of I as
$I=\dfrac{3}{32}\left( \dfrac{-\cos 2x}{2} \right)-\dfrac{1}{32}\left( \dfrac{-\cos 6x}{6} \right)+c$
$I=\dfrac{-3\cos 2x}{64}+\dfrac{\cos 6x}{192}+c$
Hence, we get $\int{{{\sin }^{3}}x{{\cos }^{3}}xdx}=\dfrac{-3\cos 2x}{64}+\dfrac{\cos 6x}{192}+c$

Note: Another approach for the problem would be that we can replace ${{\cos }^{3}}x$ and ${{\sin }^{3}}x$ by relations :-
${{\sin }^{3}}x=3\sin x-4{{\sin }^{3}}x$
${{\cos }^{3}}x=4{{\cos }^{3}}x-3\cos x$
And hence, multiply them and simplify it.
Always try to convert $\int{{{\sin }^{^{m}}}x{{\cos }^{n}}xdx}$ by $\sin 2x=2\sin x\cos x$ , if m and n are odd numbers and same, otherwise suppose ${{\tan }^{2}}x=t$ .