Question

What is the integral of ${\sin ^2}\left( x \right){\cos ^4}\left( x \right)$ ?

Answer 1

Generally, the integrals are classified into two types, definite integral and indefinite integral: a definite integral contains upper and lower limits whereas an indefinite integral does not contain upper and lower limits.
Here, we are given an indefinite integral and we are asked to calculate the value of$\int {{{\sin }^2}\left( x \right){{\cos }^4}\left( x \right)} dx$
Formula to be used:
Some formulae that we need to apply in the solution are as follows.
$\cos 2\theta = 1 - 2{\sin ^2}\theta$
$\cos 2\theta = 2{\cos ^2}\theta - 1$
$1 - {\cos ^2}\theta = {\sin ^2}2\theta$
$\int {dx} = x + C$
$\int {\cos xdx = } \sin x + C$
$\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$

Complete step by step answer:
First, let us simplify ${\sin ^2}\left( x \right){\cos ^4}\left( x \right)$
We know that $\cos 2x = 1 - 2{\sin ^2}x$
$\Rightarrow 2{\sin ^2}x = 1 - \cos 2x$
$\Rightarrow {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}$ …………$\left( 1 \right)$
Also, we have $\cos 2x = 2{\cos ^2}x - 1$
$\Rightarrow \cos 2x + 1 = 2{\cos ^2}x$
$\Rightarrow {\cos ^2}x = \dfrac{{\cos 2x + 1}}{2}$ ………..$\left( 2 \right)$
Now, we shall substitute the equations$\left( 1 \right)$and$\left( 2 \right)$in ${\sin ^2}\left( x \right){\cos ^4}\left( x \right)$
Thus, ${\sin ^2}\left( x \right){\cos ^4}\left( x \right) = \dfrac{{1 - \cos 2x}}{2}{\left( {\dfrac{{\cos 2x + 1}}{2}} \right)^2}$
$= \dfrac{1}{{2 \times 4}}\left( {1 - \cos 2x} \right)\left( {1 + \cos 2x} \right)\left( {1 + \cos 2x} \right)$
$= \dfrac{1}{8}\left( {1 - {{\cos }^2}2x} \right)\left( {1 + \cos 2x} \right)$
Now, we shall apply $1 - {\cos ^2}\theta = {\sin ^2}2\theta$.
Thus, we get ${\sin ^2}\left( x \right){\cos ^4}\left( x \right) = \dfrac{1}{8}{\sin ^2}2x\left( {1 + \cos 2x} \right)$
$= \dfrac{1}{8}{\sin ^2}2x + \dfrac{1}{8}{\sin ^2}2x\cos 2x$
Now, we shall apply the integral on both sides.
The addition rule of indefinite integral states that the sum of two functions is the sum of the So, we need to apply the addition rule.
$\int {{{\sin }^2}\left( x \right){{\cos }^4}\left( x \right)} dx = \int {\left( {\dfrac{1}{8}{{\sin }^2}2x + \dfrac{1}{8}{{\sin }^2}2x\cos 2x} \right)} dx$
$= \dfrac{1}{8}\int {{{\sin }^2}2xdx + \dfrac{1}{8}\int {{{\sin }^2}2x\cos 2x} } dx$ …………..$\left( 3 \right)$
Let us consider ${I_1} = \dfrac{1}{8}\int {{{\sin }^2}} 2xdx$ and${I_2} = \dfrac{1}{8}\int {{{\sin }^2}} 2x\cos 2xdx$, then we need to solve them separately.
${I_1} = \dfrac{1}{8}\int {{{\sin }^2}} 2xdx$
Now, we shall apply${\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}$
$\Rightarrow {I_1} = \dfrac{1}{8}\int {\dfrac{{1 - \cos 2\left( {2x} \right)}}{2}} dx$
$\Rightarrow {I_1} = \dfrac{1}{8}\int {\dfrac{{1 - \cos 4x}}{2}} dx$
$\Rightarrow {I_1} = \dfrac{1}{8} \times \dfrac{1}{2}\int {\left( {1 - \cos 4x} \right)} dx$
$\Rightarrow {I_1} = \dfrac{1}{{16}}\left[ {\int {dx - \int {\cos 4xdx} } } \right]$
Now, we need to apply the formulae$\int {dx} = x + C$ and$\int {\cos xdx = } \sin x + C$
$\Rightarrow {I_1} = \dfrac{1}{{16}}\left[ {x - \dfrac{{\sin 4x}}{4}} \right] + {C_1}$ where ${C_1}$ is the constant of integration………………..$\left( 4 \right)$
Consider${I_2} = \dfrac{1}{8}\int {{{\sin }^2}} 2x\cos 2xdx$
${I_2} = \dfrac{1}{8}\int {\left( {\cos 2x} \right){{\sin }^2}} 2xdx$
Let $t = \sin 2x$
Now, differentiate $t = \sin 2x$with respect to$x$.
$dt = 2\cos 2xdx$
$\Rightarrow \cos 2xdx = \dfrac{{dt}}{2}$
Now, ${I_2} = \dfrac{1}{8}\int {\left( {\cos 2x} \right){{\sin }^2}} 2xdx$becomes,
${I_2} = \dfrac{1}{8}\int {{t^2}\dfrac{{dt}}{2}}$
$\Rightarrow {I_2} = \dfrac{1}{8} \times \dfrac{1}{2}\int {{t^2}dt}$
$\Rightarrow {I_2} = \dfrac{1}{{16}}\int {{t^2}dt}$
$\Rightarrow {I_2} = \dfrac{1}{{16}} \times \dfrac{{{t^3}}}{3} + {C_2}$ (Here we have applied$\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$)
$\Rightarrow {I_2} = \dfrac{1}{{16}} \times \dfrac{{{{\sin }^3}2x}}{3} + {C_2}$ where${C_2}$ is the constant of integration ….$\left( 5 \right)$
Now, we shall substitute the equations$\left( 4 \right)$ and $\left( 5 \right)$in the equation$\left( 3 \right)$.
$\int {{{\sin }^2}\left( x \right){{\cos }^4}\left( x \right)} dx = \dfrac{1}{{16}}\left[ {x - \dfrac{{\sin 4x}}{4}} \right] + \dfrac{1}{{16}} \times \dfrac{{{{\sin }^3}2x}}{3} + C$ where$C$ is the constant of integration.
$= \dfrac{x}{{16}} - \dfrac{{\sin 4x}}{{64}} + \dfrac{{{{\sin }^3}2x}}{{48}} + C$
Hence, $\int {{{\sin }^2}\left( x \right){{\cos }^4}\left( x \right)} dx = \dfrac{x}{{16}} - \dfrac{{\sin 4x}}{{64}} + \dfrac{{{{\sin }^3}2x}}{{48}} + C$

Note:
We all know that differentiation is the process of finding the derivation of the functions whereas process integration is to find the antiderivative of a function and hence, these two processes are said to be inverse to each other. That means, integration is the inverse process of differentiation and also known as the anti-differentiation.