Question

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

Answer 1

Hint: To find the integral of $ {\sin ^4}\left( x \right).{\cos ^2}\left( x \right) $ , we have to write $ {\sin ^4}x $ as $ \left( {{{\sin }^2}x \cdot {{\sin }^2}x} \right) $ and then take $ {\sin ^2}\left( x \right) \cdot {\cos ^2}\left( x \right) $ as one term and $ {\sin ^2}\left( x \right) $ as one term. Now, use the formula $ {\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2} $ and $ \dfrac{{\sin 2\theta }}{2} = \sin \theta \cos \theta $ and then on simplifying the expression again use the formula $ {\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2} $ . Now, the integral is in simple form and can be easily solved.

Complete step by step solution:
In this question, we have to find the integral of $ {\sin ^4}\left( x \right).{\cos ^2}\left( x \right) $ .
 $ \int {{{\sin }^4}\left( x \right).{{\cos }^2}\left( x \right)dx} $ - - - - - - - - - (1)
Here, we can write $ {\sin ^4}x $ as $ \left( {{{\sin }^2}x \cdot {{\sin }^2}x} \right) $ . Therefore, equation (1) becomes
  $ \Rightarrow \int {{{\sin }^4}\left( x \right).{{\cos }^2}\left( x \right)dx} = \int {{{\sin }^2}\left( x \right) \cdot {{\sin }^2}\left( x \right) \cdot {{\cos }^2}\left( x \right)dx} $ - - - - - - (2)
Now, take $ {\sin ^2}\left( x \right) \cdot {\cos ^2}\left( x \right) $ as one term and $ {\sin ^2}\left( x \right) $ as one term. Therefore, equation (2) becomes
 $ \Rightarrow \int {{{\sin }^4}\left( x \right).{{\cos }^2}\left( x \right)dx} = \int {{{\sin }^2}x \cdot {{\left( {\sin x \cdot \cos x} \right)}^2}dx} $ - - - - - - - (3)
Now, here we need to use some trigonometric formulas to simplify the expression.
We know that, $ \cos 2\theta = 1 - 2{\sin ^2}\theta $ .
 $
   \Rightarrow \cos 2\theta - 1 = - 2{\sin ^2}\theta \\
   \Rightarrow {\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2} \;
  $
And $ \sin 2\theta = 2\sin \theta \cos \theta $
 $ \Rightarrow \dfrac{{\sin 2\theta }}{2} = \sin \theta \cos \theta $
Therefore, using this relations, our equation (3) becomes
 $ \Rightarrow \int {{{\sin }^4}\left( x \right).{{\cos }^2}\left( x \right)dx} = \int {\dfrac{{\left( {1 - \cos 2x} \right)}}{2} \cdot {{\left( {\dfrac{{\sin 2x}}{2}} \right)}^2}dx} $
Taking the constant terms out of the integral sign, we get
 $ \Rightarrow \int {{{\sin }^4}\left( x \right).{{\cos }^2}\left( x \right)dx} = \dfrac{1}{8}\int {\left( {1 - \cos 2x} \right){{\sin }^2}2xdx} $ - - - - - - - (4)
Here, again $ {\sin ^2}2x = \dfrac{{1 - \cos 4x}}{2} $ . Therefore, equation (4) becomes
 $ \Rightarrow \int {{{\sin }^4}\left( x \right).{{\cos }^2}\left( x \right)dx} = \dfrac{1}{8}\int {\left( {1 - \cos 2x} \right)\left( {\dfrac{{1 - \cos 4x}}{2}} \right)dx} $ - - - - - - (5)
Taking 2 out of integral sign, equation (5) becomes
 $ \Rightarrow \int {{{\sin }^4}\left( x \right).{{\cos }^2}\left( x \right)dx} = \dfrac{1}{{16}}\int {\left( {1 - \cos 2x} \right)\left( {1 - \cos 4x} \right)dx} $
Now, opening the brackets, we get
 $ \Rightarrow \int {{{\sin }^4}\left( x \right).{{\cos }^2}\left( x \right)dx} = \dfrac{1}{{16}}\int {\left( {1 - \cos 4x - \cos 2x + \cos 2x \cdot \cos 4x} \right)dx} $ - - - - - - - - (6)
Now, we have the formula $ 2\cos x\cos y = \cos \left( {x - y} \right) + \cos \left( {x + y} \right) $ .
 $ \Rightarrow \cos x\cos y = \dfrac{1}{2}\left[ {\cos \left( {x - y} \right) + \cos \left( {x + y} \right)} \right] $
Therefore, $ \cos 4x \cdot \cos 2x $ can be written as $ \dfrac{1}{2}\left[ {\cos 2x + \cos 6x} \right] $ .
Therefore, equation (6) will become
 $ \Rightarrow \int {{{\sin }^4}\left( x \right).{{\cos }^2}\left( x \right)dx} = \dfrac{1}{{16}}\int {\left( {1 - \cos 4x - \cos 2x + \dfrac{1}{2}\left( {\cos 2x + \cos 6x} \right)} \right)dx} $
Now, separating all the terms, we get
 $ \Rightarrow \int {{{\sin }^4}\left( x \right).{{\cos }^2}\left( x \right)dx} = \dfrac{1}{{16}}\left[ {\int {1 \cdot dx - \int {\cos 4xdx} } - \int {\cos 2xdx + \dfrac{1}{2}\int {\cos 2xdx + \dfrac{1}{2}\int {\cos 6xdx} } } } \right] $
\[
   = \dfrac{1}{{16}}\left[ {x - \dfrac{{\sin 4x}}{4} - \dfrac{{\sin 2x}}{2} + \dfrac{1}{2}\dfrac{{\sin 2x}}{2} + \dfrac{1}{2}\dfrac{{\sin 6x}}{6}} \right] \\
   = \dfrac{1}{{16}}\left[ {x - \dfrac{{\sin 4x}}{4} - \dfrac{{\sin 2x}}{2} + \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 6x}}{{12}}} \right] \\
   = \dfrac{1}{{16}}\left[ {x - \dfrac{{\sin 4x}}{4} - \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 6x}}{{12}}} \right] \;
 \]
This is our final answer. Therefore, $ \int {{{\sin }^4}\left( x \right).{{\cos }^2}\left( x \right)dx} = \dfrac{1}{{16}}\left[ {x - \dfrac{{\sin 4x}}{4} - \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 6x}}{{12}}} \right] $ .
So, the correct answer is “ $ \dfrac{1}{{16}}\left[ {x - \dfrac{{\sin 4x}}{4} - \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 6x}}{{12}}} \right] $ .”.

Note: We cannot directly integrate trigonometric functions with power greater than 1 as there is no direct formula for it. We have to use the relations and formulas to simplify the expression so that we can integrate it easily. So, while finding the integral of trigonometric functions with power greater than 1, always look for the relations that can be used to simplify the expression.