Question

Let $n \ge 2$ be a natural number and $0 < 	heta < \frac{\pi }{2}.$ Then $\smallint \frac{{{{({{\sin }^n}	heta - \sin 	heta )}^{\frac{1}{n}}}\cos 	heta }}{{{{\sin }^{n + 1}}	heta }}d	heta $ is equal to (where C is a constant of integration)

Accepted Answer

$\frac{n}{{{n^2} - 1}}{\left( {1 - \frac{1}{{{{\sin }^{n - 1}}	heta }}} ight)^{\frac{{n + 1}}{n}}} + C$

Answer

$\frac{n}{{{n^2} - 1}}{\left( {1 - \frac{1}{{{{\sin }^{n + 1}}	heta }}} ight)^{\frac{{n + 1}}{n}}} + C$

Answer

$\frac{n}{{{n^2} - 1}}{\left( {1 + \frac{1}{{{{\sin }^{n - 1}}	heta }}} ight)^{\frac{{n + 1}}{n}}} + C$

Answer

$\frac{n}{{{n^2} + 1}}{\left( {1 - \frac{1}{{{{\sin }^{n - 1}}	heta }}} ight)^{\frac{{n + 1}}{n}}} + C$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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