Question

What is the integral of $\int {{{\sin }^5}\left( x \right)dx}$?

Answer 1

Here in this question given an Indefinite integral, we have to find the integrated value of the given trigonometric function. This can be solved by substitution method and later integrated by using the standard trigonometric formula of integration. And by further simplification we get the required solution.

Complete step-by-step solution:
Integration is the inverse process of differentiation. An integral which is not having any upper and lower limit is known as an indefinite integral.
Consider the given function.
$\Rightarrow \,\int {{{\sin }^5}\left( x \right)dx}$
It can be rewritten as:
$\Rightarrow \,\int {{{\sin }^4}\left( x \right)\sin \left( x \right)dx}$
Or
$\Rightarrow \,\int {{{\left( {{{\sin }^2}\left( x \right)} \right)}^2}\sin \left( x \right)dx}$-------(1)
Now, using the one of the standard identity of trigonometry i.e., ${\sin ^2}\left( \theta \right) + {\cos ^2}\left( \theta \right) = 1$
On rearranging we can also written as: ${\sin ^2}\left( \theta \right) = 1 - {\cos ^2}\left( \theta \right)$
On substituting ${\sin ^2}\left( x \right)$ in equation (1), we have
$\Rightarrow \,\int {{{\left( {1 - {{\cos }^2}\left( x \right)} \right)}^2}\sin \left( x \right)dx}$---------(2)
Apply the Substitution put $u = \cos \left( x \right)$, then
On differentiating u with respect to x, we get
$\dfrac{{du}}{{dx}} = - \sin \left( x \right)$
Or it can be written as
$du = - \sin \left( x \right)dx$ or $- du = \sin \left( x \right)dx$.
Then, Integral (2) becomes
$\Rightarrow \,\int {{{\left( {1 - {u^2}} \right)}^2}\left( { - du} \right)}$
$\Rightarrow \, - \int {{{\left( {1 - {u^2}} \right)}^2}du}$-------(3)
Now, using a algebraic identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ here a=1 and b=u, then equation (3) becomes
$\Rightarrow \, - \int {\left( {{1^2} - 2\left( 1 \right)\left( {{u^2}} \right) + {{\left( {{u^2}} \right)}^2}} \right)du}$
On substitution, we have
$\Rightarrow \, - \int {\left( {1 - 2{u^2} + {u^4}} \right)du}$
Integrate each term with respect to u, then
$\Rightarrow \, - \int {du} + 2\int {{u^2}\,du} - \int {{u^4}du}$
As we know, integration of $\int {x\,dx = \dfrac{{{x^2}}}{2} + c}$, then
$\Rightarrow \,\, - u + 2\dfrac{{{u^3}}}{3} - \dfrac{{{u^5}}}{5} + C$
$\Rightarrow \,\, - u + \dfrac{2}{3}{u^3} - \dfrac{1}{5}{u^5} + C$
Substitute the u value, then
$\Rightarrow \,\, - \cos \left( x \right) + \dfrac{2}{3}{\cos ^3}\left( x \right) - \dfrac{1}{5}{\cos ^5}\left( x \right) + C$.
Where, C is an integrating constant.
Hence, it’s a required solution.

Note: By simplifying the question using the substitution we can integrate the given function easily. If we apply integration directly it may be complicated to solve further. So, simplification is needed. We must know the differentiation and integration formulas. The standard integration formulas for the trigonometric ratios must know.