# Finding the indefinite integral with rational sin

I feel like my approach thus far is valid, but I've reached a spot where it feel like things are going wrong. I've changed the $$1/\sin^2 \to \csc^2$$ and distributed it to make it easier to integrate. $$\sin^2 \times \csc^2 = 1$$, so those cancel... but that leaves me with a constant $$6$$, which if my variable were $$x$$ I'd feel right at home. But this is $$\theta$$, and making that $$6$$ into a $$6 \theta$$ just seems like I've gone wrong somewhere.

Can anyone give me a push in the right direction?

Here's my approach so far:

\eqalign { \text {Question: } & \int \frac { 7 - 6 \sin ^ 2 ( \theta ) } { \sin ^ 2 ( \theta ) } \, d \theta \\ \text {My answer: } & \int ( 7 - 6 \sin ^ 2 \theta ) ( \csc ^ 2 \theta ) \, d \theta \\ & \int ( 7 \csc ^ 2 \theta - 6 \sin ^ 2 \theta \csc ^ 2 \theta ) \, d \theta \\ & \int ( 7 \csc ^ 2 \theta - 6 ) \, d \theta \\ & 7 \cot \theta \> \cdots \hfil }

• Strictly speaking, you're not changing the $6$ into a $6\theta$; you're changing the $6\,\mathrm d\theta$ into a $6\theta$. Maybe that will make you feel better about it. Anyway, the $6\theta$ is correct; subtract that from the $-7\cot\theta$ (and tack on a ${}+C$ if you're expected to have that in your answers), and you'll be done. Aug 3, 2021 at 3:15
• Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. Aug 3, 2021 at 3:16
• I've submitted an edit with math formatting, so as soon as it gets approved by someone with enough reputation, you can open your question for editing and look at how I formatted it. See the link in José's comment above for more advice on how to do this in your questions. Aug 3, 2021 at 3:35

Yes, that $$6$$ becomes $$6\theta$$. That is$$\int7\csc^2(\theta)-6\,\mathrm d\theta=-7\cot(\theta)-6\theta+C.$$
• There's a sign error here; it's $-6$, not $+6$. Aug 3, 2021 at 3:17