Question

Evaluate the value of $\int {{{\sin }^5}xdx} $.

Answer 1

Hint:
Here, we are asked to find the value of $\int {{{\sin }^5}xdx} $.
Let $I = \int {{{\sin }^5}xdx} $ and write ${\sin ^5}x$ as ${\sin ^4}x \times \sin x$.
Then, ${\sin ^4}x$ can be written as ${\left( {{{\sin }^2}x} \right)^2}$ and ${\sin ^2}x = 1 - {\cos ^2}x$.
Thus, solve the question further to get the required answer.

Complete step by step solution:
Here, we are asked to find the value of $\int {{{\sin }^5}xdx} $ .
Let $I = \int {{{\sin }^5}xdx} $ .
Now, we can write ${\sin ^5}x$ as ${\sin ^4}x \times \sin x$ .
 $\therefore I = \int {{{\sin }^4}x \times \sin xdx} $
Also, ${\sin ^4}x$ can be written as ${\left( {{{\sin }^2}x} \right)^2}$ .
 $\therefore I = \int {{{\left( {{{\sin }^2}x} \right)}^2}\sin xdx} $
Applying the property ${\sin ^2}x = 1 - {\cos ^2}x$ in the above value of I.
 $\therefore I = \int {{{\left( {1 - {{\cos }^2}x} \right)}^2}\sin xdx} $
Now, to solve further, let $\cos x = t$
 $
  \therefore - \sin xdx = dt \\
  \therefore \sin xdx = - dt \\
 $
 $
  \therefore I = - \int {{{\left( {1 - {t^2}} \right)}^2}dt} \\
  \therefore I = - \int {\left( {1 - 2{t^2} + {t^4}} \right)dt} \\
  \therefore I = - \int {dt} + 2\int {{t^2}dt} - \int {{t^4}dt} \\
  \therefore I = - t + 2\dfrac{{{t^3}}}{3} - \dfrac{{{t^5}}}{5} + C \\
  \therefore I = - \dfrac{{{t^5}}}{5} - \dfrac{{2{t^3}}}{3} - t + C \\
 $
Now, returning back the value of t as $\cos x$ .

\[\therefore I = - \dfrac{{{{\cos }^5}x}}{5} + \dfrac{{2{{\cos }^3}x}}{3} - \cos x + C\]
Thus, $\int {{{\sin }^5}xdx} = - \dfrac{{{{\cos }^5}x}}{5} + \dfrac{{2{{\cos }^3}x}}{3} - \cos x + C$.

Note:
Some properties of definite integration:
1) $\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {\left( {a + b - f\left( x \right)} \right)dx} } $
2) $\int\limits_{ - a}^a {f\left( x \right)dx} = - \int\limits_a^{ - a} {f\left( x \right)dx} $
3) $\int\limits_a^b {kxdx = } k\int\limits_a^b {xdx} $
4) $\int\limits_a^a {f\left( x \right)dx = } 0$
 $\int\limits_a^b {f\left( x \right)dx} = - \int\limits_b^a {f\left( x \right)dx} $