Question

How do you solve \[\int {({{\sin }^4}x)({{\cos }^2}x)dx} \] ?

Answer 1

Hint: For finding a very small part of a whole quantity, we use derivative. Integration is the inverse of derivative and is also called antiderivative, so as the name suggests antiderivative means finding the whole quantity from a given small part. Thus we have to find the integration of the given quantity. Now, we have to find the integration of a trigonometric function, so we have to simplify the function by using trigonometric identities and then find its integration.

Complete step by step solution:
We have to find \[\int {({{\sin }^4}x)({{\cos }^2}x)dx} \]
Now, ${\sin ^4}x{\cos ^2}x = {\sin ^2}x.({\sin ^2}x{\cos ^2}x)$
Multiplying and dividing this equation with 4, we get –
$
   \Rightarrow {\sin ^4}x{\cos ^2}x = {\sin ^2}x.\dfrac{{4{{\sin }^2}x{{\cos }^2}x}}{4} \\
   \Rightarrow {\sin ^4}x{\cos ^2}x = {\sin ^2}x.\dfrac{{{{(2\sin x\cos x)}^2}}}{4} = {\sin ^2}x.\dfrac{{{{\sin }^2}2x}}{4} \\
 $
Now, we know that
$ \Rightarrow \cos 2x = 1 - 2{\sin ^2}x \Rightarrow {\sin ^2}x = \dfrac{1}{2}(1 - \cos 2x)$
Using this value in the above equation, we get –
$ \Rightarrow {\sin ^4}x{\cos ^2}x = \dfrac{{1 - \cos 2x}}{2}.\dfrac{{{{\sin }^2}2x}}{4} = \dfrac{1}{8}({\sin ^2}2x - \cos 2x{\sin ^2}2x)$
Multiplying the numerator and the denominator of the equation by 2, we get –
$
   \Rightarrow {\sin ^4}x{\cos ^2}x = \dfrac{1}{8}(\dfrac{1}{2} \times 2{\sin ^2}2x - \dfrac{1}{2} \times 2\cos 2x{\sin ^2}2x) \\
   \Rightarrow {\sin ^4}x{\cos ^2}x = \dfrac{1}{8}(\dfrac{1}{2}(1 - \cos 4x) - \dfrac{1}{2}\sin 4x\sin 2x) \\
 $
We know that $2\sin a\sin b = \cos (a - b) - \cos (a + b)$
Using this identity, we get –
$
   \Rightarrow {\sin ^4}x{\cos ^2}x = \dfrac{1}{{16}}(1 - \cos 4x - \dfrac{1}{2}(\cos 2x - \cos 6x)) \\
   \Rightarrow {\sin ^4}x{\cos ^2}x = \dfrac{1}{{16}}(1 - \cos 4x - \dfrac{{\cos 2x}}{2} + \dfrac{{\cos 6x}}{2}) \\
   \Rightarrow \int {{{\sin }^4}x{{\cos }^2}x} = \int {\dfrac{1}{{16}}(1 - \cos 4x - \dfrac{{\cos 2x}}{2} + \dfrac{{\cos 6x}}{2})} \\
   \Rightarrow \int {{{\sin }^4}x{{\cos }^2}x} = \dfrac{1}{{16}}\{ x - \dfrac{{\sin 4x}}{4} - (\dfrac{{\sin 2x}}{4} + \dfrac{{\sin 6x}}{{12}})\} + c \\
 $
We know that
$
   \Rightarrow \sin 3x = 3\sin x - 4{\sin ^3}x \\
   \Rightarrow \sin 6x = \sin 3(2x) = 3\sin 2x - 4{\sin ^3}2x \\
 $
Using this value in the above equation, we get –
$
   \Rightarrow \int {{{\sin }^4}x{{\cos }^2}x} = \dfrac{1}{{16}}\{ x - \dfrac{{\sin 4x}}{4} - \dfrac{{3\sin 2x + 3\sin 2x - 4{{\sin }^3}2x}}{{12}}\} + c \\
   \Rightarrow \int {{{\sin }^4}x{{\cos }^2}x} = \dfrac{1}{{16}}(x - \dfrac{{\sin 4x}}{4} - \dfrac{{{{\sin }^3}2x}}{3}) + c \\
 $

Note: In this question, we have to find the integration of the trigonometric function, so we must know the trigonometric identities and convert it in the form such that the power of the trigonometric function is 1. In differential calculus, we have to find the derivative of a function but in integration, the differentiation of a function is given and we have to find the function. An integral that is expressed with upper and lower limits is called a definite integral while an indefinite integral is expressed without limits like in this question. Varying the value of the arbitrary constant, one can get different values of integral of a function.