Determine $$\int \frac{\sec^2x-2021}{\sin^{2021}x} dx$$
I tried dividing by $\cos^{2021}x$ in both numerator and denominator which gave me a simplified form of $$\dfrac{\sec^{2021}x(\sec^2x-2021)}{\tan^{2021}x}$$ but that doesn't seem to help much.
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7$\begingroup$ What's the source of this question? The appearance of $2021$ may lead people to wonder if it's from an on-going contest. $\endgroup$– BlueJul 17, 2021 at 13:33
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1$\begingroup$ No actually,you could replace $2021$ with $n$. $\endgroup$– a_i_rJul 17, 2021 at 13:53
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3$\begingroup$ @Blue It jee advance question jeeadv.ac.in/pastqp.php see the maths section $\endgroup$– jasmineJul 17, 2021 at 13:59
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1 Answer
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Integrating by parts $$\int \frac{\sec^2(x)}{\sin^{n}(x)}\, dx=\frac{\tan(x)}{\sin^{n}(x)}-\int\frac{\tan(x)(-n\cos(x))}{\sin^{n+1}(x)}\,dx=\frac{\tan(x)}{\sin^{n}(x)}+\int\frac{n}{\sin^{n}(x)}\,dx.$$ Hence, after moving the remaining integral to the other side, we find $$\int \frac{\sec^2(x)-n}{\sin^{n}(x)}\, dx=\frac{\tan(x)}{\sin^{n}(x)}+c.$$
