Question

Solve the following expression-
$\smallint {\cos ^5}{\text{x}}.cosecxdx$

Answer 1

- Hint: Integration is a method by which we can find the area under the function of the curve. It is the reverse process of differentiation. The given expression cannot be integrated directly. We need to alter it and substitute a suitable value to make it integrable, and then proceed. We will also use some formulas as-
$\begin{align}
  &cosecx = \dfrac{1}{{sinx}}...(1) \\
  &{\cos ^2}{\text{x}} + {\sin ^2}{\text{x}} = 1...(2) \\
\end{align} $

$\begin{align}
  &\smallint \dfrac{1}{x}dx = \log x + c...\left( 3 \right) \\
  &\smallint {x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}...\left( 4 \right) \\
\end{align} $

Complete step-by-step solution -

We have been given that-
$\smallint {\cos ^5}{\text{x}}.cosecxdx$
Using property (1) we can write that-
$\begin{align}
   &= \smallint \dfrac{{{{\cos }^5}x}}{{\sin x}}dx \\
   &= \smallint \dfrac{{{{\cos }^4}x}}{{\sin x}}\cos xdx \\
   &= \smallint \dfrac{{{{\left( {{{\cos }^2}x} \right)}^2}}}{{\sin x}}\cos xdx \\
\end{align} $

Using property (2) we can write that-
$\begin{align}
  & = \smallint \dfrac{{{{\left( {1 - {{\sin }^2}{\text{x}}} \right)}^2}}}{{sinx}}cosxdx \\
 & By\;{\left( {{\text{a}} + {\text{b}}} \right)^2} = {{\text{a}}^2} + {{\text{b}}^2} + 2ab \\
  & = \smallint \left( {\dfrac{{1 + {{\sin }^4} - 2{{\sin }^2}{\text{x}}}}{{sinx}}} \right)cosxdx \\
\end{align} $

Now, let us assume sinx to be t.
t = sinx …(5)
By partially differentiating this equation we get-
dt = cosxdx …(6)
By substituting equations (5) and (6) in the expression we get-
$\begin{align}
  & = \smallint \left( {\dfrac{{1 + {{\text{t}}^4} - 2{{\text{t}}^2}}}{{\text{t}}}} \right)dt \\
  & = \smallint \left( {\dfrac{1}{{\text{t}}} - 2{\text{t}} + {{\text{t}}^3}} \right)dt \\
\end{align} $
Using properties 3 and 4,
$\begin{align}
  & = logt - \dfrac{{2{{\text{t}}^2}}}{2} + \dfrac{{{{\text{t}}^4}}}{4} + {\text{C}} \\
  & = \dfrac{{{{\sin }^4}{\text{x}}}}{4} - {\sin ^2}{\text{x}} + \log \left( {sinx} \right) + {\text{C}} \\
\end{align} $

Here C is the integration constant. This is the required answer.

Note: In such types of questions, we need to have knowledge of multiple concepts from trigonometry to integral calculus. We need to change the expression so that it can be converted into an integrable form. Also, we should always remember to add a constant of integration whenever we solve indefinite integrals.