Question

If \[{I_n} = \int\limits_0^\pi {{e^x}{{(\sin x)}^n}dx} \] then \[\dfrac{{{I_3}}}{{{I_1}}}\] is equal to
(A)\[\dfrac{3}{5}\]
(B)\[\dfrac{1}{5}\]
(C)\[1\]
(D)\[\dfrac{2}{5}\]

Answer 1

Hint: In this question, we have to choose the correct option for the particular required relation. They give the general relation, by using that we can change its general relation to the required general relation. Then we substituting the particular term to find the value to choose from the option.
In calculus, and more generally in mathematical analysis, integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and anti-derivative.

Formula used: If \[u\] and \[v\] both are derivable functions,\[\int {u\dfrac{{dv}}{{dx}}dx = uv - \int {v\dfrac{{du}}{{dx}}} } dx\]
Differentiation formula with respect to $x$
$\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$
$\dfrac{{d\left( {\sin x} \right)}}{{dx}} = \cos x$
 $\dfrac{{d\left( {\cos x} \right)}}{{dx}} = - \sin x$
$\dfrac{{dy}}{{dx}} = {e^x} \Rightarrow y = \int {{e^x}dx} $
Trigonometric relations and values,
\[\sin \pi = 0\] and \[\sin 0 = 0\]
\[{\sin ^{n - 1}}\pi = 0\] and \[{\sin ^{n - 1}}0 = 0\]
${\cos ^2}x = 1 - {\sin ^2}x$

Complete step-by-step answer:
It is given that, \[{I_n} = \int\limits_0^\pi {{e^x}{{(\sin x)}^n}dx} \]
We need to find out the value of \[\dfrac{{{I_3}}}{{{I_1}}}\].
 We have,\[{I_n} = \int\limits_0^\pi {{e^x}{{(\sin x)}^n}dx} \]
Applying by parts integration we get,
By using \[\int {u\dfrac{{dv}}{{dx}}dx = uv - \int {v\dfrac{{du}}{{dx}}} } dx\] equating the R.H.S of this equation to given relation we get,
\[{I_n} = \int\limits_0^\pi {{e^x}{{(\sin x)}^n}dx} = \int {u\dfrac{{dv}}{{dx}}dx} \] from this,
$u = {(\sin x)^n}$ and $\dfrac{{du}}{{dx}} = \dfrac{{d{{(\sin x)}^n}}}{{dx}}$,
$\dfrac{{dv}}{{dx}} = {e^x}$ and $v = \int\limits_0^\pi {{e^x}dx} $
Substituting the values of $u$, $\dfrac{{du}}{{dx}}$, $v$ and $\dfrac{{dv}}{{dx}}$ in \[\int {u\dfrac{{dv}}{{dx}}dx = uv - \int {v\dfrac{{du}}{{dx}}} } dx\] we get,
\[ \Rightarrow {I_n} = {(\sin x)^n}\int\limits_0^\pi {{e^x}dx} - \int\limits_0^\pi {\dfrac{{d{{(\sin x)}^n}}}{{dx}}} {e^x}dx\]
Let us consider the term,
\[\dfrac{{d{{(\sin x)}^n}}}{{dx}}\] Differentiating with respect to \[x\],
Differentiate using the differentiating formula,
$\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$
$\dfrac{{d\left( {\sin x} \right)}}{{dx}} = \cos x$
 $\dfrac{{d\left( {\cos x} \right)}}{{dx}} = - \sin x$
\[ \Rightarrow \dfrac{{d{{(\sin x)}^n}}}{{dx}} = n{\sin ^{n - 1}}x\left( {\cos x} \right)\]
Substituting the term \[\dfrac{{d{{(\sin x)}^n}}}{{dx}} = n{\sin ^{n - 1}}x\left( {\cos x} \right)\] and integrating we get,
\[ \Rightarrow {I_n} = \left[ {{{(\sin x)}^n}{e^x}} \right]\begin{array}{*{20}{c}}
  \pi \\
  0
\end{array} - \int\limits_0^\pi {n{{\sin }^{n - 1}}x\cos x} {e^x}dx\]
Applying the limit values for the integral term,
\[ \Rightarrow {I_n} = \left[ {{{(\sin \pi )}^n}{e^\pi } - {{(\sin 0)}^n}{e^0}} \right] - \int\limits_0^\pi {n{{\sin }^{n - 1}}x\cos x} {e^x}dx\]
\[\sin \pi = 0\] and \[\sin 0 = 0\] by using this we get,
\[ \Rightarrow {I_n} = (0 - 0) - n\int\limits_0^\pi {{{\sin }^{n - 1}}x\cos x} {e^x}dx\]
\[ \Rightarrow {I_n} = - n\int\limits_0^\pi {{{\sin }^{n - 1}}x\cos x} {e^x}dx\]
Now we will take \[{\sin ^{n - 1}}x\cos x\] as first term and \[{e^x}\]as second term and applying integral by parts we get,
\[ \Rightarrow {I_n} = - n\int\limits_0^\pi {{{\sin }^{n - 1}}x\cos x} {e^x}dx = \int {u\dfrac{{dv}}{{dx}}dx} \] from this,
\[u = - n{\sin ^{n - 1}}x\cos x\] and $\dfrac{{du}}{{dx}} = \dfrac{{ - nd\left( {{{\sin }^{n - 1}}x\cos x} \right)}}{{dx}}$,
$\dfrac{{dv}}{{dx}} = {e^x}$ and $v = \int\limits_0^\pi {{e^x}dx} $
By Substituting the values of $u$, $\dfrac{{du}}{{dx}}$, $v$ and $\dfrac{{dv}}{{dx}}$ in \[\int {u\dfrac{{dv}}{{dx}}dx = uv - \int {v\dfrac{{du}}{{dx}}} } dx\] we get,
\[ \Rightarrow {I_n} = - n{\sin ^{n - 1}}x\cos x\int\limits_0^\pi {{e^x}dx} - \left( { - n} \right)\int\limits_0^\pi {\dfrac{{d({{\sin }^{n - 1}}x\cos x)}}{{dx}}{e^x}dx} \]
Integrating the terms,
\[ \Rightarrow {I_n} = - n\left[ {{{\sin }^{n - 1}}x\cos x.{e^x}} \right]\begin{array}{*{20}{c}}
  \pi \\
  0
\end{array} + n\int\limits_0^\pi {\dfrac{{d({{\sin }^{n - 1}}x\cos x)}}{{dx}}{e^x}dx} \]
Applying the limit values for the integral term,
\[ \Rightarrow {I_n} = - n\left[ {{{\sin }^{n - 1}}\pi \cos \pi .{e^\pi } - {{\sin }^{n - 1}}0\cos 0.{e^0}} \right] + n\int\limits_0^\pi {\dfrac{{d({{\sin }^{n - 1}}x\cos x)}}{{dx}}{e^x}dx} \]
\[{\sin ^{n - 1}}\pi = 0\] and \[{\sin ^{n - 1}}0 = 0\] by using this we get,

\[ \Rightarrow {I_n} = (0 - 0) + n\int\limits_0^\pi {\dfrac{{d({{\sin }^{n - 1}}x\cos x)}}{{dx}}{e^x}dx} \]
\[ \Rightarrow {I_n} = n\int\limits_0^\pi {\dfrac{{d({{\sin }^{n - 1}}x\cos x)}}{{dx}}{e^x}dx} \]
Let us consider the term,
\[\dfrac{{d\left( {{{\sin }^{n - 1}}x\cos x} \right)}}{{dx}}\] Differentiating with respect to \[x\],
Differentiate using the differentiating formula,
$\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$
$\dfrac{{d\left( {\sin x} \right)}}{{dx}} = \cos x$
 $\dfrac{{d\left( {\cos x} \right)}}{{dx}} = - \sin x$
\[ \Rightarrow \dfrac{{d\left( {{{\sin }^{n - 1}}x\cos x} \right)}}{{dx}} = (n - 1){\sin ^{n - 2}}x\cos x.\cos x - {\sin ^{n - 1}}x\sin x\]
Multiplying the common term $\cos x$,
\[ \Rightarrow (n - 1){\sin ^{n - 2}}x{\cos ^2}x - {\sin ^{n - 1}}x\sin x\]
We know that ${\cos ^2}x = 1 - {\sin ^2}x$,
\[ \Rightarrow (n - 1){\sin ^{n - 2}}x(1 - {\sin ^2}x) - {\sin ^{n - 1}}x\sin x\]
Multiply this \[(n - 1){\sin ^{n - 2}}x\] into \[(1 - {\sin ^2}x)\] and simplifying the power values of \[{\sin ^{n - 1}}x\sin x\] we get,
\[ \Rightarrow (n - 1){\sin ^{n - 2}}x - (n - 1){\sin ^n}x - {\sin ^n}x\]
Multiplying \[(n - 1)\] into \[{\sin ^n}x\],
\[ \Rightarrow (n - 1){\sin ^{n - 2}}x - n{\sin ^n}x + {\sin ^n}x - {\sin ^n}x\]
Simplifying we get,
\[ \Rightarrow (n - 1){\sin ^{n - 2}}x - n{\sin ^n}x\]
Thus we get,
\[\dfrac{{d\left( {{{\sin }^{n - 1}}x\cos x} \right)}}{{dx}} = (n - 1){\sin ^{n - 2}}x - n{\sin ^n}x\]
Substituting the value of \[\dfrac{{d\left( {{{\sin }^{n - 1}}x\cos x} \right)}}{{dx}}\] in \[{I_n} = n\int\limits_0^\pi {\dfrac{{d({{\sin }^{n - 1}}x\cos x)}}{{dx}}{e^x}dx} \],
\[ \Rightarrow {I_n} = n\int\limits_0^\pi {\left[ {(n - 1){{\sin }^{n - 2}}x - n{{\sin }^n}x} \right]} {e^x}dx\]
$n\; {{\& }} n - 1$ are independent from $x$, taking out them from integral,
\[ \Rightarrow {I_n} = n(n - 1)\int\limits_0^\pi {{{\sin }^{n - 2}}x.{e^x}dx - {n^2}\int\limits_0^\pi {{{\sin }^n}x} } {e^x}dx\]
Since, \[{I_n} = \int\limits_0^\pi {{e^x}{{(\sin x)}^n}dx} = \int\limits_0^\pi {{{\sin }^n}x.{e^x}dx} \]
For $n = n - 2$,
\[{I_{n - 2}} = \int\limits_0^\pi {{e^x}{{(\sin x)}^{n - 2}}dx} = \int\limits_0^\pi {{{\sin }^{n - 2}}x.{e^x}dx} \]
By using this we can rewrite,
\[ \Rightarrow {I_n} = n(n - 1){I_{n - 2}} - {n^2}{I_n}\]
Simplifying we get,
\[ \Rightarrow {I_n} + {n^2}{I_n} = n(n - 1){I_{n - 2}}\]
Taking out the common term \[{I_n}\] in the L.H.S,
\[ \Rightarrow {I_n}(1 + {n^2}) = n(n - 1){I_{n - 2}}\]
Arranging the \[{I_n}\] terms in L.H.S,
\[ \Rightarrow \dfrac{{{I_n}}}{{{I_{n - 2}}}} = \dfrac{{n(n - 1)}}{{1 + {n^2}}}\]
Putting, \[n = 3\] we get,
\[ \Rightarrow \dfrac{{{I_3}}}{{{I_{3 - 2}}}} = \dfrac{{{I_3}}}{{{I_1}}} = \dfrac{{3(3 - 1)}}{{1 + {3^2}}}\]
Simplifying we get,
\[ \Rightarrow \dfrac{{3 \times 2}}{{1 + 9}} = \dfrac{6}{{10}} = \dfrac{3}{5}\]
\[\therefore \dfrac{{{I_3}}}{{{I_1}}} = \dfrac{3}{5}\]

So, the correct answer is “Option A”.

Note: Integration by parts is a technique for performing indefinite integration \[\int {udv} \] or definite integration \[\int\limits_a^b {udv} \]
 by expanding the differential of a product of functions \[d(uv)\] and expressing the original integral in terms of a known integral \[\int {vdu} \]. A single integration by parts starts with
\[d(uv) = udv + vdu\]
And integrating both sides,
\[\int {d(uv) = uv} = \int {udv} + \int {vdu} \]
Rearranging we get,
\[\int {udv} = uv - \int {vdu} \]