0
$\begingroup$

This is the problem I want done. However my teacher challenged the class to solve it without using substitution. I understand up to the 3rd equal sign. How would I proceed from there, without using substitution? \begin{align*} \int \frac{1}{1+\sin x} \, dx &= \int \frac{1 - \sin x}{1 - \sin^2 x} \, dx \\ &= \int \frac{1 - \sin x}{\cos^2 x} \, dx \\ &= \int \sec^2 x - \frac{\sin x}{\cos^2 x} \, dx \\ &= \tan x - \int \frac{-du}{u^2}, \quad u = \cos x, du = -\sin x \, dx \\ &= \tan x - \frac{1}{u} + C \\ &= \tan x - \sec x + C. \end{align*}

$\endgroup$
2
$\begingroup$

Note that: $\frac{\sin x}{\cos^2 x} = \tan x\sec x$ and the derivative of $\sec x$ is $\tan x\sec x$.

$\endgroup$
  • $\begingroup$ Wow, I feel like an idiot. Thanks for pointing me to the obvious... $\endgroup$ – AfronPie Sep 20 '16 at 23:51

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.