Question

If ${I_n} = \int\limits_0^\pi {{e^x}(\sin x} {)^n}dx$ then, find the value of $\dfrac{{{I_3}}}{{{I_1}}}$

Answer 1

Hint: As given in the question to find the value of $\dfrac{{{I_3}}}{{{I_1}}}$ first of all we have to solve the given integration. As we can see that given integration is product of two functions/terms so, to solve the given integration we have to use the formula to find the product of two functions/terms (or by part method) but, before using the formula we have to decide which function/term we should choose as first function/term and second function/term and to choose the first function/term we have to use the LIATE rule which can explained as:
L – Logs
I – Inverse
A – Algebraic
T – Trig
E – Exponential
As given in the question there are two types of functions ${e^x}$(exponential function) and $\sin x$(trig function) so, according to the LIATE rule our first function will be $\sin x$and second function will be ${e^x}$.

Formula used: $\int {f(x)g(x)dx = f(x)\int {g(x)dx - \int {\left( {\dfrac{{df(x)}}{{dx}}\int {g(x)dx} } \right)d(x)........................} } } (1)$
Where,
$f(x)$is the first function/term and,
$g(x)$is the second function/term.
$\int {{a^n}dx = n{a^{n - 1}}} .................................(2)$
$\int {{e^x}dx = {e^x} + c.................................(3)} $
$\dfrac{{d(\sin x)}}{{dx}} = \cos x...............................(4)$
$\dfrac{{d(\cos x)}}{{dx}} = - \sin x$
$\dfrac{{d(\sin x)}}{{dx}} = \cos x$

Complete step-by-step answer:
Given,
${I_n} = \int\limits_0^\pi {{e^x}(\sin x} {)^n}dx$
Step 1: According to the question given in the question first of all we will use the LIATE rule to choose our first and second function/term and according to it $\sin x$is our first function and second function is ${e^x}$.
Step 2: Now, to solve the given integration we will use formula (1) which is used to find the integration of two products by using by part rule.
${I_n} = {(\sin x)^n}\int\limits_0^\pi {{e^x}} dx - \int\limits_0^\pi {\left( {\dfrac{{d{{(\sin x)}^n}}}{{dx}}\int {{e^x}dx} } \right)} dx$
Step 3: Now, to solve the integration we will use the formulas (2, 3 and 4) as mentioned above in the solution hint section.
${I_n} = \left[ {{{\left( {\sin x} \right)}^n}{e^x}} \right]_0^\pi - \int\limits_0^\pi {\left[ {n{{(\sin x)}^{n - 1}}(\cos x){e^x}} \right]} dx$
Step 4: On placing the value of x in the equation obtained just above,
${I_n} = \left[ {{{\left( {\sin \pi } \right)}^n}{e^\pi } - {{(\sin 0)}^n}{e^0}} \right] - n\int\limits_0^\pi {\left[ {{{(\sin x)}^{n - 1}}(\cos x){e^x}} \right]} dx$
As we know $\sin {0^0} = 1$and ${e^0} = 1,\sin \pi = 0,$hence,
${I_n} = \left[ {\left( 0 \right) \times 1 - 0 \times 1} \right] - n\int\limits_0^\pi {\left[ {{{\left( {\sin x} \right)}^{n - 1}}(\cos x){e^x}} \right]dx} $
Hence,
${I_n} = - n\int\limits_0^\pi {\left[ {{{\left( {\sin x} \right)}^{n - 1}}(\cos x){e^x}} \right]dx} $…………………………………….(5)
Step 5: As we can see that in the obtained equation (5) we have to use the formula (2) again to solve the integration and now our first term will be ${(\sin x)^{n - 1}}(\cos x)$and our second term will be ${e^x}$
Hence,
${I_n} = - n{(\sin x)^{n - 1}}\cos x\int\limits_0^\pi {{e^x}dx - n\int\limits_0^\pi {\left[ {\dfrac{{d{{(\sin x)}^{n - 1}}\cos x}}{{dx}}\int {{e^x}} } \right]dx} } $…………………………………….(6)
Now to solve the differentiation of ${(\sin x)^{n - 1}}(\cos x)$ we have to use the formula given below:
\[\dfrac{{df(x)g(x)}}{{dx}} = f(x)\dfrac{{dg(x)}}{{dx}} + g(x)\dfrac{{df(x)}}{{dx}}\]………………………………………(7)
Where $f(x)$is the first term and $g(x)$is the second term.
Step 6: On differentiating${(\sin x)^{n - 1}}(\cos x)$with respect to x, with the help of the formula (7)
$ = \dfrac{{d\left[ {{{\left( {\sin x} \right)}^{n - 1}}\cos x} \right]}}{{dx}}$
$ = \cos x\dfrac{{d{{(\sin x)}^{n - 1}}}}{{dx}} + {(\sin x)^{n - 1}}\dfrac{{d(\cos x)}}{{dx}}$
Now to find the differentiation of the equation obtained just above we have to use the formulas mentioned in the solution hint.
$
   = \cos x(n - 1){(\sin x)^{n - 1 - 1}}\cos x + {(\sin x)^{n - 1}}( - \sin x) \\
   = (n - 1){(\sin x)^{n - 2}}{(\cos x)^2} - {(\sin x)^{n - 1 + 1}} \\
   = (n - 1){(\sin x)^{n - 2}}{(\cos x)^2} - {(\sin x)^n}.................................(8) \\
 $
As we know,
${(\cos x)^2} = 1 - {(\sin x)^2}$
On substituting in equation (8),
$
   = (n - 1){(\sin x)^{n - 2}}(1 - {\sin ^2}x) - {(\sin x)^n} \\
   = (n - 1){(\sin x)^{n - 2}} - (n - 1){(\sin x)^{n - 2 + 2}} - {(\sin x)^n} \\
   = (n - 1){(\sin x)^{n - 2}} - (n - 1){(\sin x)^n} - {(\sin x)^n} \\
   = (n - 1){(\sin x)^{n - 2}} - n{(\sin x)^n} + {(\sin x)^n} - {(\sin x)^n} \\
 $
As the equation obtained above $ + {(\sin x)^n} - {(\sin x)^n}$will be eliminated hence, the obtained equation is:
$ = (n - 1){(\sin x)^{n - 2}} - n{(\sin x)^n}$ ………………………………(9)
Step 7: On substituting the obtained equation (9) in equation (6),
$
  {I_n} = - n{(\sin x)^{n - 1}}\cos x\int\limits_0^\pi {{e^x}dx + n\int\limits_0^\pi {\left[ {\{ (n - 1){{(\sin x)}^{n - 2}} - n{{(\sin x)}^n}\} \int {{e^x}dx} } \right]} } dx \\
  {I_n} = [ - n{(\sin x)^{n - 1}}(\cos x){e^x}]_0^\pi + n\int\limits_0^\pi {\left[ {(n - 1){{(\sin x)}^{n - 2}}{e^x} - n{{(\sin x)}^n}{e^x}} \right]} dx \\
 $
On solving the obtained equation,
${I_n} = \left[ { - n{{(\sin \pi )}^{n - 1}}(\cos \pi ){e^\pi } - n{{(\sin 0)}^{n - 1}}(\cos 0){e^0}} \right] + n\int\limits_0^\pi {[(n - 1){{(\sin x)}^{n - 2}}{e^x}]dx - \int\limits_0^\pi n [{{(\sin x)}^n}} {e^x}]dx$As we know,
$
  \sin \pi = 0, \\
  \cos \pi = - 1, \\
  \sin {0^0} = 0, \\
  \cos {0^0} = 1, \\
  {e^{}} = 1 \\
 $
Hence,
$
  {I_n} = [ - n{(0)^{n - 1}}( - 1) \times 1 - n{(0)^{n - 1}}(1) \times 1 + n\int\limits_0^\pi {[(n - 1){{(\sin x)}^{n - 2}}{e^x}]dx - \int\limits_0^\pi {[n{{(\sin x)}^n}{e^x}]dx} } \\
  {I_n} = n\int\limits_0^\pi {[(n - 1){{(\sin x)}^{n - 2}}{e^x}]dx - \int\limits_0^\pi {[n{{(\sin x)}^n}{e^x}]dx} } ...............................(10) \\
 $
Step 8: As given in the question: ${I_n} = \int\limits_0^\pi {{e^x}{{(\sin x)}^n}dx} $. Hence, on substituting the value of ${I_n}$in the obtained equation (10)
$
  {I_n} = n[(n - 1){I_{n - 2}} - n{I_n}] \\
  {I_n} = n(n - 1){I_{n - 2}} - {n^2}{I_n} \\
  {I_n}(1 + {n^2}) = n(n - 1){I_{n - 2}} \\
 $
On cross-multiplication the equation obtained above,
$\dfrac{{{I_n}}}{{{I_{n - 1}}}} = \dfrac{{n(n - 1)}}{{(1 + {n^2})}}$…………………………………(11)
Step 9: On placing the value n = 3 in the equation (11)
$\dfrac{{{I_3}}}{{{I_{3 - 2}}}} = \dfrac{{3(3 - 1)}}{{(1 + {3^2})}}$
Hence, on solving the obtained equation,
$
  \dfrac{{{I_3}}}{{{I_1}}} = \dfrac{6}{{10}} \\
  \dfrac{{{I_3}}}{{{I_1}}} = \dfrac{3}{5} \\
 $
Final solution: Hence, we have obtained the value of $\dfrac{{{I_3}}}{{{I_1}}}$ which is: $\dfrac{3}{5}$.

Note: If the given integration is a product of two functions/terms then to solve this type of integration we will have to use the Integration by part rule. Where we have to choose the first and second function/term using the LIATE rule where, L belongs to Logs, I belongs to Inverse, A belongs to Algebraic, T belongs to Trig, and E belongs to Exponential.
If the equation is getting complicated try two solutions in parts.
The constant expression, allows us to ignore the constant coefficient in the expression while we integrate the rest of the expression.