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.

  • 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. $\endgroup$
    – Blue
    Jul 17, 2021 at 13:33
  • 1
    $\begingroup$ No actually,you could replace $2021$ with $n$. $\endgroup$
    – a_i_r
    Jul 17, 2021 at 13:53
  • 3
    $\begingroup$ @Blue It jee advance question jeeadv.ac.in/pastqp.php see the maths section $\endgroup$
    – jasmine
    Jul 17, 2021 at 13:59

1 Answer 1


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 $$\int \frac{\sec^2(x)-n}{\sin^{n}(x)}\, dx=\frac{\tan(x)}{\sin^{n}(x)}+c.$$


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