Question

Find the indefinite integral of:
[1 + 4cotx]/[4 - cotx] dx

I would like a hint please

Answer 1

I supposed
∫ csc^4 x cot^6 x dx =
- ∫ 1/sin^2 x cot^6 x d(cot x) =
- ∫ (sin^2 x + cos^2 x)/sin^2 x * cot^6 x d(cot x) =
- ∫ (1 + cot^2 x) cot^6 x d(cot x) =
- ∫ ( cot^6 x + cot^8 x) d(cot x) =
-(1/7)cot^7 (x) - (1/9)cot^8 (x) + C

Or
-(1/7)(cot (x))^7 - (1/9)(cot (x))^9 + C

Answer 2

Trig sub,
\[\text{Let} ~u=\tan \left(\frac{x}{2}\right)\] ,\[and~~du=\frac{1}{2} \sec ^2\left(\frac{x}{2}\right) d x\]

Answer 3

You'll get
\[\huge \int\limits \frac{2 \left(1-\frac{2 \left(u^2-1\right)}{u}\right)}{\left(u^2+1\right) \left(\frac{u^2-1}{2 u}+4\right)} \, du\]
Simplifying
\[\text{}=\int\limits \frac{-8 u^2+4 u+8}{u^4+8 u^3+8 u-1} \, du\]
Seaparate using partial fractions,
\[\text{}=\int\limits \left(-\frac{2 u}{u^2+1}+\frac{1}{u-\sqrt{17}+4}+\frac{1}{u+\sqrt{17}+4}\right) \, du\]
Solve

Answer 4

You can check your final answer with
\[\ln \left(\tan \left(\frac{x}{2}\right)-\sqrt{17}+4\right)+\ln \left(\tan \left(\frac{x}{2}\right)+\sqrt{17}+4\right)-\ln \left(\sec ^2\left(\frac{x}{2}\right)\right)+c\]

Answer 5

How were you able to use tanx? All I see is cotx...