Verified Solution • 2025–26 Session moderate

QJEEM19Ma7423: JEE Main 2019 Mathematics - Class 12 JEEM19Ma7423 Solution: Let n 2 be a natural number and Then is equal to - (where C is…

Let n ≥\ge 2 be a natural number and 0 < \theta < {\pi \over 2}. Then ∫(sin⁡nθ−sin⁡θ)1/ncos⁡θsin⁡n+1θ dθ\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta is equal to - (where C is a constant of integration)

Multiple Choice Options

  • A nn2−1(1+1sin⁡n−1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 + {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C
  • B nn2−1(1−1sin⁡n+1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n + 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C
  • C nn2−1(1−1sin⁡n−1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C
  • D nn2+1(1−1sin⁡n−1θ)n+1n+C{n \over {{n^2} + 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C

Step-by-Step Verified Solution

∫(sin⁡nθ−sin⁡θ)1/ncos⁡θsin⁡n+1θ dθ\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta

=∫sin⁡θ(1−1sin⁡n−1θ)1/nsin⁡n+1θ dθ= \int {{{\sin \theta {{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)}^{1/n}}} \over {{{\sin }^{n + 1}}\theta }}} \,d\theta

Put 1−1sin⁡n−1θ=t1 - {1 \over {{{\sin }^{n - 1}}\theta }} = t

So (n−1)sin⁡nθcos⁡θdθ=dt{{\left( {n - 1} \right)} \over {{{\sin }^n}\theta }}\cos \theta d\theta = dt

Now   1n−1∫(t)1n+1dt{1 \over {n - 1}}\int {{{\left( t \right)}^{{1 \over n} + 1}}dt}

=1(n−1)(t)1n+11n+1+C= {1 \over {\left( {n - 1} \right)}}{{{{\left( t \right)}^{{1 \over n} + 1}}} \over {{1 \over n} + 1}} + C

=n(n2−1)(1−1sin⁡n−1θ)1n+1+C= {n \over {\left( {n^2 - 1} \right)}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{1 \over n} + 1}} + C

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