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: A
Solution :
| \[\int_{{}}^{{}}{\frac{{{\left( {{\sin }^{n}}\theta -\sin \theta \right)}^{\frac{1}{n}}}\cos \theta }{{{\sin }^{n+1}}\theta }}d\theta \] |
| \[\int_{{}}^{{}}{\frac{{{\left( {{t}^{n}}-t \right)}^{n}}dt}{{{t}^{n+1}}}}\] (put\[\sin \theta =t\]) |
| \[\int_{{}}^{{}}{\frac{t{{\left( 1-\frac{1}{{{t}^{n-1}}} \right)}^{\frac{1}{n}}}}{{{t}^{n+1}}}}dt=\int_{{}}^{{}}{\frac{{{\left( 1-\frac{1}{{{t}^{n-1}}} \right)}^{\frac{1}{n}}}}{{{t}^{n}}}}dt\] |
| Put, \[1-\frac{1}{{{t}^{n-1}}}=z\] |
| \[\Rightarrow \] \[\frac{(n-1)}{{{t}^{n}}}dt=dz,\] |
| \[\Rightarrow \] \[I=\frac{1}{n-1}\int_{{}}^{{}}{{{z}^{\frac{1}{n}}}dz}\]\[=\frac{{{z}^{\frac{1}{n}+1}}}{\left( \frac{1}{n}+1 \right)(n-1)}+c=\frac{n{{(1-{{t}^{1-n}})}^{n+1}}}{{{n}^{2}}-1}+c\] |
| \[=\frac{n}{{{n}^{2}}-1}{{\left( 1-\frac{1}{{{\sin }^{n-1}}\theta } \right)}^{\frac{n+1}{n}}}+c.\] |
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