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Indefinite Integrals question

2019 · 10 Jan · Shift 1 · Q26
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  5. /2019 · 10 Jan · Shift 1 · Q26

Indefinite Integrals question

2019 · 10 Jan · Shift 1 · Q26

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
Let n ≥\ge≥ 2 be a natural number and 0<θ<π2.0 \lt \theta \lt {\pi \over 2}.0<θ<2π​. 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∫sinn+1θ(sinnθ−sinθ)1/ncosθ​dθ is equal to - (where C is a constant of integration)
  1. 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}}} + Cn2−1n​(1+sinn−1θ1​)nn+1​+C
  2. 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}}} + Cn2−1n​(1−sinn+1θ1​)nn+1​+C
  3. 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}}} + Cn2−1n​(1−sinn−1θ1​)nn+1​+C
  4. 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}}} + Cn2+1n​(1−sinn−1θ1​)nn+1​+C
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