Question

Solve the given integral, \[\int{{{\sin }^{4}}x\cdot dx}\]

Answer 1

Hint: Split the integral as a square function. Apply double angle identity for \[{{\sin }^{2}}x\] and simplify it. Find again the double angle identity for \[{{\cos }^{2}}x\] . Thus, solve the expression formed by using basic integration formulas.

Complete step-by-step answer:
We have been given the integral which we need to solve. Let us take it as I. Thus, we can write it as,
\[I=\int{{{\sin }^{4}}x\cdot dx}\] ………………………………(1)
This integral is mostly about rewriting the functions. As a rule of thumb, if the power is even, we can use the double angle formula.
The double angle formula says that,
\[\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x\] …………………………….(2)
Now, we know that \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]
\[\therefore {{\cos }^{2}}x=1-{{\sin }^{2}}x\] , putting this value in (2), we get
\[\cos 2x=1-{{\sin }^{2}}x-{{\sin }^{2}}x\]
\[\cos 2x=1-2{{\sin }^{2}}x\]
Hence, we get that \[{{\sin }^{2}}x=\dfrac{1}{2}\left( 1-\cos 2x \right)dx\] ……………………………(3)
Now, we can write equation (1) as,
\[I=\int{{{\sin }^{2}}x\cdot }{{\sin }^{2}}xdx\]
Now put \[{{\sin }^{2}}x=\dfrac{1}{2}\left( 1-\cos 2x \right)\] in the above expression.
\[\therefore I=\int{\dfrac{1}{2}\left( 1-\cos 2x \right)\cdot \dfrac{1}{2}\left( 1-\cos 2x \right)dx}\]
\[=\dfrac{1}{4}{{\left( 1-\cos 2x \right)}^{2}}dx\]
We know the identity, \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\] . Split the above as,
\[\therefore I=\dfrac{1}{4}\int{\left[ 1-2\cos 2x+{{\cos }^{2}}2x \right]dx}\] ……………………………………..(4)
Now from (2), \[\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x\]
Put \[{{\sin }^{2}}x=1-{{\cos }^{2}}x\]
\[\therefore \cos 2x={{\cos }^{2}}x-\left( 1-{{\cos }^{2}}x \right)\]
\[\cos 2x={{\cos }^{2}}x-1+{{\cos }^{2}}x\]
\[\therefore 2{{\cos }^{2}}x=\cos 2x+1\]
\[\therefore {{\cos }^{2}}x=\dfrac{1}{2}\left( 1+\cos 2x \right)\]
\[\therefore {{\cos }^{2}}\left( 2x \right)=\dfrac{1}{2}\left( 1+\cos \left( 2\left( 2x \right) \right) \right)=\dfrac{1}{2}\left( 1+4x \right)\] …………………………..(5)
Substitute the values of (5) in (4).
\[I=\dfrac{1}{4}\int{\left[ 1-2\cos 2x+{{\cos }^{2}}2x \right]}dx\]
\[=\dfrac{1}{4}\int{\left[ 1-2\cos 2x+\dfrac{1}{2}\left( 1+\cos 4x \right) \right]}dx\]
\[=\dfrac{1}{4}\int{\left[ 1-2\cos 2x+\dfrac{1}{2}+\dfrac{\cos 4x}{2} \right]}dx\]
\[I=\dfrac{1}{4}\left( \int{1dx}-2\int{\cos 2x\text{ }dx}+\int{\dfrac{1}{2}\cdot dx}+\dfrac{1}{2}\int{\cos 4x}\cdot dx \right)\] ………………………………….(6)
Let us first see the integration of \[\int{2\cos 2x\text{ }dx}\] and \[\int{\cos 4x\text{ }dx}\] .
Let us take \[{{I}_{1}}=\int{2\cos 2x\text{ }dx}\]
Let us put \[u=2x\] in the above expression,
\[\therefore \] derivative of u, \[du=2\cdot dx\]
\[\therefore dx=\dfrac{1}{2}du\]
\[\therefore {{I}_{1}}=\int{2\cos 2x\text{ }dx}\]
\[=\int{\dfrac{2\cos u}{2}\cdot du}=\int{\cos u}\cdot du\]
We know that \[\int{\cos u\cdot du}=\sin u\]
\[\therefore {{I}_{1}}=\sin u=\sin 2x\]
\[\Rightarrow {{I}_{1}}=\sin 2x\] ……………………………………(7)
Let us take \[u=4x\]
\[\therefore \] derivative of \[u\cdot du=4dx\]
\[\therefore dx=\dfrac{1}{4}du\]
\[\therefore {{I}_{2}}=\int{\cos 4x\text{ }dx}=\dfrac{1}{4}\int{\cos u\cdot du}=\dfrac{1}{4}\sin u\]
\[\therefore \int{\cos 4xdx}=\dfrac{\sin 4x}{4}\] …………………………………(8)
Now put the value of (7) and (8) in (6), we get
\[I=\dfrac{1}{4}\left[ \int{1dx-\int{2\cos 2x\text{ }dx+\int{\dfrac{1}{2}dx}}}+\dfrac{1}{2}\int{\cos 4x\text{ }dx} \right]\]
\[I=\dfrac{1}{4}\left[ x-\sin 2x+\dfrac{x}{2}+\dfrac{1}{2}\left( \dfrac{1}{4}\sin 4x \right) \right]+c\]
\[I=\dfrac{1}{4}\left[ x-\sin 2x+\dfrac{x}{2}+\dfrac{1}{8}\sin 4x \right]+c\]
\[=\dfrac{x}{4}-\dfrac{\sin 2x}{4}+\dfrac{x}{8}+\dfrac{\sin 4x}{32}+c\]
\[=\left( \dfrac{x}{4}+\dfrac{x}{8} \right)-\dfrac{\sin 2x}{4}+\dfrac{\sin 4x}{32}+c\]
\[\therefore I=\dfrac{3x}{6}-\dfrac{1}{4}\sin \left( 2x \right)+\dfrac{\sin 4x}{32}+c\]
Hence, we got the required solution as,
\[\int{{{\sin }^{4}}xdx}=\dfrac{3x}{6}-\dfrac{\sin 2x}{4}+\dfrac{\sin 4x}{32}+c\]

Note: The basics of solution of this integral is that you know the double angle formulas. There are double angle and half angle identities for sine, cosine and tangent functions. It is important that you remember the identities.