Question

Find $\int {{{\sin }^4}} x \cdot dx$.

Answer 1

Hint: Here we need to find the integration of the given function. We will first simplify the function by expanding the function. We will use the algebraic identities to expand the given function and then we will use the trigonometric identities in the expansion. After using the identities and mathematical operations, we will integrate each term and then combine the terms to get the required answer.

Complete step by step solution:
Let the value of integration of the given function be $I$.
$ I = \int {{{\sin }^4}} x \cdot dx$
We know from the basic trigonometric identities that ${\sin ^2}\theta = \dfrac{1}{2}\left( {1 - \cos 2\theta } \right)$.
Using this trigonometric identity in the above equation, we get
$ \Rightarrow I = \int {\dfrac{1}{2}\left( {1 - \cos 2x} \right) \times \dfrac{1}{2}\left( {1 - \cos 2x} \right)} dx$
Now, we will multiply the required terms inside the bracket.
$ \Rightarrow I = \int {\dfrac{1}{4}\left( {1 + {{\cos }^2}2x - 2\cos 2x} \right)} dx$
On further simplifying the terms, we get
$ \Rightarrow I = \dfrac{1}{4}\left[ {\int 1 \cdot dx + \int {{{\cos }^2}2x \cdot dx} - 2\int {\cos 2x \cdot dx} } \right]$
We know from the basic trigonometric identities that:-
${\cos ^2}\theta = \dfrac{1}{2}\left( {1 + \cos 2\theta } \right)$
Using this trigonometric identity here, we get
$ \Rightarrow I = \dfrac{1}{4}\left[ {\int 1 \cdot dx + \int {\dfrac{1}{2}\left( {1 + \cos 4x} \right) \cdot dx} - 2\int {\cos 2x \cdot dx} } \right]$
On further simplifying the terms, we get
$ \Rightarrow I = \dfrac{1}{4}\left[ {\int 1 \cdot dx + \dfrac{1}{2}\int {\cos 4x \cdot dx} + \dfrac{1}{2}\int {1 \cdot dx} - 2\int {\cos 2x \cdot dx} } \right]$
Now, we will integrate all the functions individually here.
$ \Rightarrow I = \dfrac{1}{4}\left[ {x + \dfrac{1}{2} \times \dfrac{1}{4} \times \sin 4x + \dfrac{1}{2} \times x - 2 \times \dfrac{1}{2} \times \sin 2x} \right]$
On multiplying the terms, we get
$ \Rightarrow I = \dfrac{1}{4}\left[ {x + \dfrac{{\sin 4x}}{8} + \dfrac{x}{2} - \sin 2x} \right]$
On adding the like terms inside the bracket, we get
$ \Rightarrow I = \dfrac{1}{4}\left[ {\dfrac{{3x}}{2} + \dfrac{{\sin 4x}}{8} - \sin 2x} \right]$
On multiplying the terms using the distributive properties of multiplication, we get
$ \Rightarrow I = \dfrac{{3x}}{8} + \dfrac{{\sin 4x}}{{32}} - \dfrac{{\sin 2x}}{4} + c$
Hence, this is the required integration of the given function.

Note:
In this question, we have used trigonometry. Trigonometry is a branch of mathematics which helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by cartographers (to make maps). It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life. The three most common trigonometric functions are the tangent function, the sine, and the cosine function. In simple terms, they are written as ‘sin’, ‘cos’, and ‘tan’.