Let $n\ge 2$ be a natural number and $0<\theta <\dfrac{\pi }{2}$. Then find the value of indefinite integral $\int{\dfrac{{{\left( {{\sin }^{n}}\theta -\sin \theta \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $? (Where C is constant of integration).
(a) $\dfrac{n}{{{n}^{2}}-1}{{\left( 1-\dfrac{1}{{{\sin }^{n+1}}\theta } \right)}^{\dfrac{n+1}{n}}}+C$,
(b) $\dfrac{1}{{{n}^{2}}+1}{{\left( 1-\dfrac{1}{{{\sin }^{n+1}}\theta } \right)}^{\dfrac{n+1}{n}}}+C$,
(c) $\dfrac{1}{n-1}{{\left( 1-\dfrac{1}{{{\sin }^{n+1}}\theta } \right)}^{\dfrac{n+1}{n}}}+C$,
(d) $\dfrac{n}{{{n}^{2}}-1}{{\left( 1+\dfrac{1}{{{\sin }^{n+1}}\theta } \right)}^{\dfrac{n+1}{n}}}+C$.
Answer
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Hint: We start solving the problem by assigning a variable to the given indefinite integral $\int{\dfrac{{{\left( {{\sin }^{n}}\theta -\sin \theta \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $. We take ${{\sin }^{n}}\theta $ common from inside of the bracket present in the numerator. We then make subsequent calculations and we then take $1-\dfrac{1}{{{\sin }^{n-1}}\theta }=t$ to convert $d\theta $ in terms of $dt$. We then use the property $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$ to solve the indefinite integral and we later substitute the value of t in the obtained results. We then make subsequent arrangements to get the required result.
Complete step-by-step solution:
According to the problem, we have to find the value of the indefinite integral $\int{\dfrac{{{\left( {{\sin }^{n}}\theta -\sin \theta \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $ given that $n\ge 2$ be a natural number and $0<\theta <\dfrac{\pi }{2}$.
Let us assume the indefinite integral be ‘I’. So, we have $I=\int{\dfrac{{{\left( {{\sin }^{n}}\theta -\sin \theta \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $.
$\Rightarrow I=\int{\dfrac{{{\left( {{\sin }^{n}}\theta \left( 1-\dfrac{\sin \theta }{{{\sin }^{n}}\theta } \right) \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $.
$\Rightarrow I=\int{\dfrac{{{\left( {{\sin }^{n}}\theta \right)}^{\dfrac{1}{n}}}{{\left( 1-\dfrac{\sin \theta }{{{\sin }^{n}}\theta } \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $.
$\Rightarrow I=\int{\dfrac{\left( \sin \theta \right){{\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $.
$\Rightarrow I=\int{\dfrac{{{\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n}}\theta }}d\theta $.
$\Rightarrow I=\int{{{\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)}^{\dfrac{1}{n}}}\left( \dfrac{\cos \theta }{{{\sin }^{n}}\theta } \right)}d\theta $ ---(1).
Let us assume $1-\dfrac{1}{{{\sin }^{n-1}}\theta }=t$ ---(2).
Let us differentiate them on both sides.
$\Rightarrow d\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)=dt$.
We know that $d\left( a+b \right)=d\left( a \right)+d\left( b \right)$.
$\Rightarrow d\left( 1 \right)-d\left( \dfrac{1}{{{\sin }^{n-1}}\theta } \right)=dt$.
We know that $d\left( a \right)=0$ and $d\left( \dfrac{1}{{{y}^{n}}} \right)=\dfrac{-n}{{{y}^{n+1}}}dy$.
$\Rightarrow 0-\left( \dfrac{-\left( n-1 \right)}{{{\sin }^{n-1+1}}\theta } \right)d\left( \sin \theta \right)=dt$.
We know that $d\left( \sin \theta \right)=\cos \theta d\theta $.
$\Rightarrow \left( \dfrac{\left( n-1 \right)}{{{\sin }^{n}}\theta } \right)\left( \cos \theta \right)d\theta =dt$.
$\Rightarrow \left( \dfrac{\left( \cos \theta \right)}{{{\sin }^{n}}\theta } \right)d\theta =\dfrac{dt}{\left( n-1 \right)}$ ---(3).
We substitute equation (2) and (3) in equation (1).
$\Rightarrow I=\int{{{\left( t \right)}^{\dfrac{1}{n}}}}\dfrac{dt}{\left( n-1 \right)}$.
$\Rightarrow I=\dfrac{1}{\left( n-1 \right)}\int{{{\left( t \right)}^{\dfrac{1}{n}}}}dt$.
We know that $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$.
$\Rightarrow I=\dfrac{1}{\left( n-1 \right)}\left( \dfrac{{{t}^{\dfrac{1}{n}+1}}}{\dfrac{1}{n}+1} \right)+C$.
$\Rightarrow I=\dfrac{1}{\left( n-1 \right)}\left( \dfrac{{{t}^{\dfrac{1}{n}+1}}}{\dfrac{1+n}{n}} \right)+C$.
$\Rightarrow I=\dfrac{n}{\left( n-1 \right)\left( n+1 \right)}\left( {{t}^{\dfrac{1+n}{n}}} \right)+C$.
$\Rightarrow I=\dfrac{n}{\left( {{n}^{2}}-1 \right)}\left( {{t}^{\dfrac{n+1}{n}}} \right)+C$ -----(4).
From equation (2) we have $t=1-\dfrac{1}{{{\sin }^{n-1}}\theta }$. We substitute this in equation (4).
\[\Rightarrow I=\dfrac{n}{\left( {{n}^{2}}-1 \right)}\left( {{\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)}^{\dfrac{n+1}{n}}} \right)+C\]. Where C is integration constant.
We have found the value of the indefinite integral $\int{\dfrac{{{\left( {{\sin }^{n}}\theta -\sin \theta \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $ as \[\dfrac{n}{\left( {{n}^{2}}-1 \right)}\left( {{\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)}^{\dfrac{n+1}{n}}} \right)+C\].
$\therefore$ If $n\ge 2$ be a natural number and $0<\theta <\dfrac{\pi }{2}$, then the value of indefinite integral $\int{\dfrac{{{\left( {{\sin }^{n}}\theta -\sin \theta \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $ is \[\dfrac{n}{\left( {{n}^{2}}-1 \right)}\left( {{\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)}^{\dfrac{n+1}{n}}} \right)+C\].
The correct option for the given problem is (a).
Note: We should not make mistakes while making calculations related to integrals. We have given an indefinite integral in the given problem, which means we have to re substitute the value of assumed variables again after getting results. If we have definite integral with limits, then there is no need to worry about re-substitution again as the values of limits change after making a change in substitution. Adding integration constant is necessary while solving an indefinite integral.
Complete step-by-step solution:
According to the problem, we have to find the value of the indefinite integral $\int{\dfrac{{{\left( {{\sin }^{n}}\theta -\sin \theta \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $ given that $n\ge 2$ be a natural number and $0<\theta <\dfrac{\pi }{2}$.
Let us assume the indefinite integral be ‘I’. So, we have $I=\int{\dfrac{{{\left( {{\sin }^{n}}\theta -\sin \theta \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $.
$\Rightarrow I=\int{\dfrac{{{\left( {{\sin }^{n}}\theta \left( 1-\dfrac{\sin \theta }{{{\sin }^{n}}\theta } \right) \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $.
$\Rightarrow I=\int{\dfrac{{{\left( {{\sin }^{n}}\theta \right)}^{\dfrac{1}{n}}}{{\left( 1-\dfrac{\sin \theta }{{{\sin }^{n}}\theta } \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $.
$\Rightarrow I=\int{\dfrac{\left( \sin \theta \right){{\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $.
$\Rightarrow I=\int{\dfrac{{{\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n}}\theta }}d\theta $.
$\Rightarrow I=\int{{{\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)}^{\dfrac{1}{n}}}\left( \dfrac{\cos \theta }{{{\sin }^{n}}\theta } \right)}d\theta $ ---(1).
Let us assume $1-\dfrac{1}{{{\sin }^{n-1}}\theta }=t$ ---(2).
Let us differentiate them on both sides.
$\Rightarrow d\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)=dt$.
We know that $d\left( a+b \right)=d\left( a \right)+d\left( b \right)$.
$\Rightarrow d\left( 1 \right)-d\left( \dfrac{1}{{{\sin }^{n-1}}\theta } \right)=dt$.
We know that $d\left( a \right)=0$ and $d\left( \dfrac{1}{{{y}^{n}}} \right)=\dfrac{-n}{{{y}^{n+1}}}dy$.
$\Rightarrow 0-\left( \dfrac{-\left( n-1 \right)}{{{\sin }^{n-1+1}}\theta } \right)d\left( \sin \theta \right)=dt$.
We know that $d\left( \sin \theta \right)=\cos \theta d\theta $.
$\Rightarrow \left( \dfrac{\left( n-1 \right)}{{{\sin }^{n}}\theta } \right)\left( \cos \theta \right)d\theta =dt$.
$\Rightarrow \left( \dfrac{\left( \cos \theta \right)}{{{\sin }^{n}}\theta } \right)d\theta =\dfrac{dt}{\left( n-1 \right)}$ ---(3).
We substitute equation (2) and (3) in equation (1).
$\Rightarrow I=\int{{{\left( t \right)}^{\dfrac{1}{n}}}}\dfrac{dt}{\left( n-1 \right)}$.
$\Rightarrow I=\dfrac{1}{\left( n-1 \right)}\int{{{\left( t \right)}^{\dfrac{1}{n}}}}dt$.
We know that $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$.
$\Rightarrow I=\dfrac{1}{\left( n-1 \right)}\left( \dfrac{{{t}^{\dfrac{1}{n}+1}}}{\dfrac{1}{n}+1} \right)+C$.
$\Rightarrow I=\dfrac{1}{\left( n-1 \right)}\left( \dfrac{{{t}^{\dfrac{1}{n}+1}}}{\dfrac{1+n}{n}} \right)+C$.
$\Rightarrow I=\dfrac{n}{\left( n-1 \right)\left( n+1 \right)}\left( {{t}^{\dfrac{1+n}{n}}} \right)+C$.
$\Rightarrow I=\dfrac{n}{\left( {{n}^{2}}-1 \right)}\left( {{t}^{\dfrac{n+1}{n}}} \right)+C$ -----(4).
From equation (2) we have $t=1-\dfrac{1}{{{\sin }^{n-1}}\theta }$. We substitute this in equation (4).
\[\Rightarrow I=\dfrac{n}{\left( {{n}^{2}}-1 \right)}\left( {{\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)}^{\dfrac{n+1}{n}}} \right)+C\]. Where C is integration constant.
We have found the value of the indefinite integral $\int{\dfrac{{{\left( {{\sin }^{n}}\theta -\sin \theta \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $ as \[\dfrac{n}{\left( {{n}^{2}}-1 \right)}\left( {{\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)}^{\dfrac{n+1}{n}}} \right)+C\].
$\therefore$ If $n\ge 2$ be a natural number and $0<\theta <\dfrac{\pi }{2}$, then the value of indefinite integral $\int{\dfrac{{{\left( {{\sin }^{n}}\theta -\sin \theta \right)}^{\dfrac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta $ is \[\dfrac{n}{\left( {{n}^{2}}-1 \right)}\left( {{\left( 1-\dfrac{1}{{{\sin }^{n-1}}\theta } \right)}^{\dfrac{n+1}{n}}} \right)+C\].
The correct option for the given problem is (a).
Note: We should not make mistakes while making calculations related to integrals. We have given an indefinite integral in the given problem, which means we have to re substitute the value of assumed variables again after getting results. If we have definite integral with limits, then there is no need to worry about re-substitution again as the values of limits change after making a change in substitution. Adding integration constant is necessary while solving an indefinite integral.
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