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

If In=∫0πex(sin⁡x)ndx{I_n} = \int\limits_0^\pi {{e^x}(\sin x} {)^n}dx then, find the value of I3I1\dfrac{{{I_3}}}{{{I_1}}}

Explanation

Solution

As given in the question to find the value of I3I1\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 ex{e^x}(exponential function) and sin⁡x\sin x(trig function) so, according to the LIATE rule our first function will be sin⁡x\sin xand second function will be ex{e^x}.

Formula used: ∫f(x)g(x)dx=f(x)∫g(x)dx−∫(df(x)dx∫g(x)dx)d(x)........................(1)\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)f(x)is the first function/term and,
g(x)g(x)is the second function/term.
∫andx=nan−1.................................(2)\int {{a^n}dx = n{a^{n - 1}}} .................................(2)
∫exdx=ex+c.................................(3)\int {{e^x}dx = {e^x} + c.................................(3)}
d(sin⁡x)dx=cos⁡x...............................(4)\dfrac{{d(\sin x)}}{{dx}} = \cos x...............................(4)
d(cos⁡x)dx=−sin⁡x\dfrac{{d(\cos x)}}{{dx}} = - \sin x
d(sin⁡x)dx=cos⁡x\dfrac{{d(\sin x)}}{{dx}} = \cos x

Complete step-by-step answer:
Given,
In=∫0πex(sin⁡x)ndx{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\sin xis our first function and second function is ex{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.
In=(sin⁡x)n∫0πexdx−∫0π(d(sin⁡x)ndx∫exdx)dx{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.
In=[(sin⁡x)nex]0π−∫0π[n(sin⁡x)n−1(cos⁡x)ex]dx{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,
In=[(sin⁡π)neπ−(sin⁡0)ne0]−n∫0π[(sin⁡x)n−1(cos⁡x)ex]dx{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⁡00=1\sin {0^0} = 1and e0=1,sin⁡π=0,{e^0} = 1,\sin \pi = 0,hence,
In=[(0)×1−0×1]−n∫0π[(sin⁡x)n−1(cos⁡x)ex]dx{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,
In=−n∫0π[(sin⁡x)n−1(cos⁡x)ex]dx{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){(\sin x)^{n - 1}}(\cos x)and our second term will be ex{e^x}
Hence,
In=−n(sin⁡x)n−1cos⁡x∫0πexdx−n∫0π[d(sin⁡x)n−1cos⁡xdx∫ex]dx{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){(\sin x)^{n - 1}}(\cos x) we have to use the formula given below:
df(x)g(x)dx=f(x)dg(x)dx+g(x)df(x)dx\dfrac{{df(x)g(x)}}{{dx}} = f(x)\dfrac{{dg(x)}}{{dx}} + g(x)\dfrac{{df(x)}}{{dx}}………………………………………(7)
Where f(x)f(x)is the first term and g(x)g(x)is the second term.
Step 6: On differentiating(sin⁡x)n−1(cos⁡x){(\sin x)^{n - 1}}(\cos x)with respect to x, with the help of the formula (7)
=d[(sin⁡x)n−1cos⁡x]dx= \dfrac{{d\left[ {{{\left( {\sin x} \right)}^{n - 1}}\cos x} \right]}}{{dx}}
=cos⁡xd(sin⁡x)n−1dx+(sin⁡x)n−1d(cos⁡x)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−1−1cos⁡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)  = \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{(\cos x)^2} = 1 - {(\sin x)^2}
On substituting in equation (8),
=(n−1)(sin⁡x)n−2(1−sin⁡2x)−(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  = (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 + {(\sin x)^n} - {(\sin x)^n}will be eliminated hence, the obtained equation is:
=(n−1)(sin⁡x)n−2−n(sin⁡x)n= (n - 1){(\sin x)^{n - 2}} - n{(\sin x)^n} ………………………………(9)
Step 7: On substituting the obtained equation (9) in equation (6),
In=−n(sin⁡x)n−1cos⁡x∫0πexdx+n∫0π[(n−1)(sin⁡x)n−2−n(sin⁡x)n∫exdx]dx In=[−n(sin⁡x)n−1(cos⁡x)ex]0π+n∫0π[(n−1)(sin⁡x)n−2ex−n(sin⁡x)nex]dx  {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,
In=[−n(sin⁡π)n−1(cos⁡π)eπ−n(sin⁡0)n−1(cos⁡0)e0]+n∫0π[(n−1)(sin⁡x)n−2ex]dx−∫0πn[(sin⁡x)nex]dx{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}]dxAs we know,
sin⁡π=0, cos⁡π=−1, sin⁡00=0, cos⁡00=1, e=1  \sin \pi = 0, \\\ \cos \pi = - 1, \\\ \sin {0^0} = 0, \\\ \cos {0^0} = 1, \\\ {e^{}} = 1 \\\
Hence,
In=[−n(0)n−1(−1)×1−n(0)n−1(1)×1+n∫0π[(n−1)(sin⁡x)n−2ex]dx−∫0π[n(sin⁡x)nex]dx In=n∫0π[(n−1)(sin⁡x)n−2ex]dx−∫0π[n(sin⁡x)nex]dx...............................(10)  {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: In=∫0πex(sin⁡x)ndx{I_n} = \int\limits_0^\pi {{e^x}{{(\sin x)}^n}dx} . Hence, on substituting the value of In{I_n}in the obtained equation (10)
In=n[(n−1)In−2−nIn] In=n(n−1)In−2−n2In In(1+n2)=n(n−1)In−2  {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,
InIn−1=n(n−1)(1+n2)\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)
I3I3−2=3(3−1)(1+32)\dfrac{{{I_3}}}{{{I_{3 - 2}}}} = \dfrac{{3(3 - 1)}}{{(1 + {3^2})}}
Hence, on solving the obtained equation,
I3I1=610 I3I1=35  \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 I3I1\dfrac{{{I_3}}}{{{I_1}}} which is: 35\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.