A) \[\frac{n}{{{n}^{2}}-1}\,{{\left( 1+\frac{1}{\sin {{\,}^{n-1}}\,\theta } \right)}^{\frac{n+1}{n}}}\,+C\]
B) \[\frac{n}{{{n}^{2}}-1}\,{{\left( 1-\frac{1}{\sin {{\,}^{n+1}}\,\theta } \right)}^{\frac{n+1}{n}}}\,+C\]
C) \[\frac{n}{{{n}^{2}}-1}\,{{\left( 1-\frac{1}{\sin {{\,}^{n-1}}\,\theta } \right)}^{\frac{n+1}{n}}}\,+C\]
D) \[\frac{n}{{{n}^{2}}+1}\,{{\left( 1-\frac{1}{\sin {{\,}^{n-1}}\,\theta } \right)}^{\frac{n+1}{n}}}\,+C\]
Correct Answer: C
Solution :
put sin \[\theta \] \[=t\Rightarrow \,\,cos\,\theta d\,\theta =dt\] \[I=\int{\frac{{{({{t}^{n}}-t)}^{1/n}}}{{{t}^{n+1}}}dt}\,\,=\,\,\int{\frac{{{\left( 1-\frac{1}{{{t}^{n-1}}} \right)}^{1/n}}\,dt}{{{t}^{n}}}}\] \[=\,\,\int{{{x}^{1/n}}\frac{dx}{(n-1)}\,\,\,\,\,\,\,\,\,\,\,\,\,put\,1-\frac{1}{{{t}^{n-1}}}\,=x}\] \[=\frac{1}{(n-1)}\frac{{{x}^{\frac{1}{n}+1}}}{\frac{1}{n+1}}+C\,\,\,\Rightarrow \,\,(n-1){{t}^{-(n-1)-1}}\,dt\,=\,dx\] \[\Rightarrow \,\,\,(n-1)\frac{dt}{{{t}^{n}}}\,\,=\,\,dx\] \[=\,\,\,\,\,\,\frac{n}{{{n}^{2}}-1}{{\left( 1-\frac{1}{{{\sin }^{n-1}}\theta } \right)}^{\frac{n+1}{n}}}+C\]You need to login to perform this action.
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