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Question: solve the given integration \( {\text{If }}{{\text{I}}_n} = \smallint {\text{co}}{{\text{s}}^n}x...

solve the given integration If In=∫cosnxdx then In= A. −1ncosn−1x sinx + (n−1n)In−2 B. cosn−1xsin⁡x+(n−1n)In−2 C.−1ncosn−1x sinx - (n−1n)In−2 D. 1ncosn−1x sinx + (n−1n)In−2  {\text{If }}{{\text{I}}_n} = \smallint {\text{co}}{{\text{s}}^n}xdx{\text{ then }}{{\text{I}}_n} = \\\ {\text{A}}{\text{. }}\dfrac{{ - 1}}{{\text{n}}}{\text{co}}{{\text{s}}^{n - 1}}x{\text{ sinx + }}\left( {\dfrac{{{\text{n}} - 1}}{{\text{n}}}} \right){{\text{I}}_{n - 2}} \\\ {\text{B}}{\text{. co}}{{\text{s}}^{n - 1}}x\sin x + \left( {\dfrac{{{\text{n}} - 1}}{{\text{n}}}} \right){{\text{I}}_{n - 2}} \\\ {\text{C}}{\text{.}}\dfrac{{ - 1}}{{\text{n}}}{\text{co}}{{\text{s}}^{n - 1}}x{\text{ sinx - }}\left( {\dfrac{{{\text{n}} - 1}}{{\text{n}}}} \right){{\text{I}}_{n - 2}} \\\ {\text{D}}{\text{. }}\dfrac{1}{{\text{n}}}{\text{co}}{{\text{s}}^{n - 1}}x{\text{ sinx + }}\left( {\dfrac{{{\text{n}} - 1}}{{\text{n}}}} \right){{\text{I}}_{n - 2}} \\\

Explanation

Solution

Hint: In this question, first we will break the term cosnx{\text{co}}{{\text{s}}^n}x into two parts ‘cosx’ and ‘cosn−1x{\text{co}}{{\text{s}}^{n - 1}}x. After this, we will apply the ‘By parts’ rule of integration to further solve. We will assume ‘cos x’ as second function and ‘cosn−1x{\text{co}}{{\text{s}}^{n - 1}}x’ as first function and then solve it further to get the required expression.

Complete step-by-step solution -
It is given that In=∫cosnxdx{{\text{I}}_n} = \smallint {\text{co}}{{\text{s}}^n}xdx . (1)
Let us first divide the function cosnx{\text{co}}{{\text{s}}^n}x into two parts ‘cosx’ and ‘cosn−1x{\text{co}}{{\text{s}}^{n - 1}}x’.
Here we will use the ‘By parts’ rule of integration.
Let ‘cosx’ as second function and ‘cosn−1x{\text{co}}{{\text{s}}^{n - 1}}x’ as first function.
We know that if ‘u’ and ‘v’ are two functions then using ‘By parts’ rule, we can write:
∫uvdx = u∫vdx - ∫(dudx×∫v dx)\smallint {\text{uvdx = u}}\smallint {\text{vdx - }}\smallint (\dfrac{{{\text{du}}}} {{dx}} \times \smallint {\text{v dx)}} , where
‘u’ is taken as first function and
‘v’ is taken as a second function.

Here we are taking ‘cosx’ as second function and cosn−1x{\text{co}}{{\text{s}}^{n - 1}}x as first function
Therefore, equation 1 can be written as:
In=∫cos⁡xcosn−1xdx=cosn−1x×∫cosx dx - ∫(d(cosn−1x)dx×∫cosx dx){{\text{I}}_n} = \smallint \cos x{\text{co}}{{\text{s}}^{n - 1}}xdx = {\text{co}}{{\text{s}}^{n - 1}}x \times \smallint {\text{cosx dx - }}\smallint \left( {\dfrac{{{\text{d(co}}{{\text{s}}^{n - 1}}x)}}{{{\text{dx}}}} \times \smallint {\text{cosx dx}}} \right)
=cosn−1x×sinx - ∫((n - 1)cosn−2x×−(sinx)×sinx dx){\text{co}}{{\text{s}}^{n - 1}}x \times {\text{sinx - }}\smallint \left( {({\text{n - 1)co}}{{\text{s}}^{n - 2}}x \times - {\text{(sinx)}} \times {\text{sinx dx}}} \right)
={\text{co}}{{\text{s}}^{n - 1}}x \times {\text{sinx - }}\left\\{ { - \smallint \left( {({\text{n - 1)co}}{{\text{s}}^{n - 2}}x \times {\text{si}}{{\text{n}}^2}{\text{x dx}}} \right)} \right\\}
=cosn−1x×sinx + ∫((n - 1)cosn−2x×sin2x dx){\text{co}}{{\text{s}}^{n - 1}}x \times {\text{sinx + }}\smallint \left( {({\text{n - 1)co}}{{\text{s}}^{n - 2}}x \times {\text{si}}{{\text{n}}^2}{\text{x dx}}} \right)

We know that sin2x=1−cos2x{\text{si}}{{\text{n}}^2}x = 1 - {\text{co}}{{\text{s}}^2}x.

    $\therefore $ ${{\text{I}}_n}$ =${\text{co}}{{\text{s}}^{n - 1}}x \times {\text{sinx + }}\smallint \left\\{ {({\text{n - 1)co}}{{\text{s}}^{n - 2}}x(1 - {{\cos }^2}{\text{x) dx}}} \right\\}$  
                   =${\text{co}}{{\text{s}}^{n - 1}}x \times {\text{sinx + (n - 1)}}\smallint \left( {{\text{co}}{{\text{s}}^{n - 2}}x\ {\text{dx}}} \right) + \left\\{ { - ({\text{n - 1)}}\smallint {\text{co}}{{\text{s}}^{n - 2}}x \times {\text{co}}{{\text{s}}^2}{\text{x dx}}} \right\\}$   
                   =${\text{co}}{{\text{s}}^{n - 1}}x \times {\text{sinx + (n - 1)}}\smallint \left( {{\text{co}}{{\text{s}}^{n - 2}}x\ {\text{dx}}} \right) - ({\text{n - 1)}}\smallint {\text{co}}{{\text{s}}^n}x\ {\text{dx}}$.

But we know that In=∫cosnx dx{{\text{I}}_n} = \smallint {\text{co}}{{\text{s}}^n}x\ {\text{dx}} .
∴In=cosn−1x×sinx + (n - 1)∫(cosn−2x dx)−(n - 1)In\therefore {{\text{I}}_n} = {\text{co}}{{\text{s}}^{n - 1}}x \times {\text{sinx + (n - 1)}}\smallint \left( {{\text{co}}{{\text{s}}^{n - 2}}x\ {\text{dx}}} \right) - ({\text{n - 1)}}{{\text{I}}_n}
Also, In−2=∫cosn−2x dx{{\text{I}}_{n - 2}} = \smallint {\text{co}}{{\text{s}}^{n - 2}}x\ {\text{dx}}.
∴In+(n - 1)In=cosn−1x×sinx + (n - 1)In−2\therefore {{\text{I}}_n} + ({\text{n - 1)}}{{\text{I}}_n} = {\text{co}}{{\text{s}}^{n - 1}}x \times {\text{sinx + (n - 1)}}{{\text{I}}_{n - 2}}
\Rightarrow {{\text{I}}_n}\left\\{ {1 + ({\text{n - 1)}}} \right\\} = {\text{co}}{{\text{s}}^{n - 1}}x \times {\text{sinx + (n - 1)}}{{\text{I}}_{n - 2}}
⇒In×n=cosn−1x×sinx + (n - 1)In−2\Rightarrow {{\text{I}}_n} \times {\text{n}} = {\text{co}}{{\text{s}}^{n - 1}}x \times {\text{sinx + (n - 1)}}{{\text{I}}_{n - 2}}
⇒In=cosn−1x×sinx + (n - 1)In−2n=cosn−1x×sinxn+(n - 1)nIn−2\Rightarrow {{\text{I}}_n} = \dfrac{{{\text{co}}{{\text{s}}^{n - 1}}x \times {\text{sinx + (n - 1)}}{{\text{I}}_{n - 2}}}}{{\text{n}}} = \dfrac{{{\text{co}}{{\text{s}}^{n - 1}}x \times {\text{sinx}}}}{{\text{n}}} + \dfrac{{({\text{n - 1)}}}}{{\text{n}}}{{\text{I}}_{n - 2}}
So, option D is correct.

Note: In this question, the important step is division of a given function into two parts. You should remember the method of solving integration using the ‘By part’ rule. Here the first and second function are generally chosen according to ILATE rule, where:
I- Inverse function
L- Logarithmic function
A- Algebraic function
T- Trigonometric function
E- Exponential function
Second function is chosen in decreasing order from top to bottom.