Let n ≥ 2 be a natural number and \(0 < \theta < \frac{\pi }{2}.{\rm{\;Then\;}}\smallint \left( {\frac{{{{\left( {{\rm{si}}{{\rm{n}}^n}\theta - {\rm{sin}}\theta } \right)}^{\frac{1}{n}}}{\rm{cos}}\theta }}{{{\rm{si}}{{\rm{n}}^{n + 1}}\theta }}} \right)d\theta\) is equal to:(Where C is a constant of integration)

From question, the INTEGRATION given is:\(I = \smallint \left( {\FRAC{{{{\left( {{\rm{si}}{{\rm{n}}^n}\;\theta - {\rm{sin\;}}\theta } \right)}^{\frac{1}{n}}}{\rm{COS\;}}\theta }}{{{\rm{si}}{{\rm{n}}^{n + 1}}\theta }}} \right)d\theta\) On putting, sin θ = t⇒ cos θ dθ = dtNow,\(\RIGHTARROW I = \smallint \frac{{{{\left( {{t^n} - t} \right)}^{1/n}}}}{{{t^{n + 1}}}}DT\) \(\Rightarrow I = \smallint \frac{{{{\left[ {{t^n}\left( {1 - \frac{t}{{{t^n}}}} \right)} \right]}^{1/n}}}}{{{t^{n + 1}}}}dt\) \(\Rightarrow I = \smallint \frac{{t{{\left( {1 - \frac{1}{{{t^{n - 1}}}}} \right)}^{1/n}}}}{{{t^{n + 1}}}}dt\) \(\Rightarrow I = \smallint \frac{{{{\left( {1 - 1/{t^{n - 1}}} \right)}^{1/n}}}}{{{t^n}}}dt\) On putting,\(\Rightarrow 1 - \frac{1}{{{t^{n - 1}}}} = u\) ∴ 1 – t-(n-1) = u\(\Rightarrow \frac{{\left( {n - 1} \right)}}{{{t^n}}}dt = du\) \(\therefore \frac{{dt}}{{{t^n}}} = \frac{{du}}{{n - 1}}\) On substituting in integral,\(\Rightarrow I = \smallint \frac{{{u^{1/n}}du}}{{n - 1}} = \frac{{{u^{\frac{1}{n} + 1}}}}{{\left( {n - 1} \right)\left( {\frac{1}{n} + 1} \right)}} + C\) \(\Rightarrow I = \frac{{n{{\left( {1 - \frac{1}{{{t^{n - 1}}}}} \right)}^{n + 1}}}}{{\left( {n - 1} \right)\left( {n + 1} \right)}} + C\) \(\left[ {u = 1 - \frac{1}{{{t^{n - 1}}}}{\rm{\;and\;}}t = {\rm{sin}}\theta } \right]\) \(\Rightarrow I=\frac{n{{\left( 1-\frac{1}{\text{si}{{\text{n}}^{n-1}}\theta } \right)}^{\frac{n+1}{n}}}}{{{n}^{2}}-1}+C\) \(\therefore I=\frac{n}{{{n}^{2}}-1}{{\left( 1-\frac{1}{\text{si}{{\text{n}}^{n-1}}\theta } \right)}^{\frac{n+1}{n}}}+C\)