JEE Main 201910 Jan, Morning shiftMathematicsMediumIntegral Calculus369× since 2002Let n ≥\ge≥ 2 be a natural number and 0<θ<π2.0 < \theta < {\pi \over 2}.0<θ<2π. Then ∫(sinnθ−sinθ)1/ncosθsinn+1θ dθ\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta∫sinn+1θ(sinnθ−sinθ)1/ncosθdθ is equal to - (where C is a constant of integration)AOption A. nn2−1(1+1sinn−1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 + {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + Cn2−1n(1+sinn−1θ1)nn+1+CBOption B. nn2−1(1−1sinn+1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n + 1}}\theta }}} \right)^{{{n + 1} \over n}}} + Cn2−1n(1−sinn+1θ1)nn+1+CCOption C. nn2−1(1−1sinn−1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + Cn2−1n(1−sinn−1θ1)nn+1+CDOption D. nn2+1(1−1sinn−1θ)n+1n+C{n \over {{n^2} + 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + Cn2+1n(1−sinn−1θ1)nn+1+CCheckSkip · show solution
