3
$\begingroup$

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.

$\endgroup$
3
  • 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

4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.