Question

integrate as soon as possible​

Answer 1

Step-by-step explanation:

$\sf \int \frac{d\theta}{4 {cos}^{2}\theta + 9{sin}^{2}\theta} \\ \\ \sf Dividing \: by \: {cos}^{2}\theta \\ \\ \sf \int \frac{{sec}^{2}\theta \: d\theta}{4 + 9{tan}^{2}\theta} \\ \\ \sf Substituting\: 3tan\theta = t \\ \\ \sf {sec}^{2}\theta \: d\theta = \frac{1}{3}dt \\ \\ \sf \int \frac{1}{3} \frac{dt}{4 + {t}^{2}} \\ \\ \sf \int \frac{1}{3}\frac {dt}{{2}^{2} + {t}^{2}} \\ \\ \sf \frac{1}{6} {tan}^{-1}\frac{t}{2} \\ \\ \sf \frac{1}{6} {tan}^{-1}\frac{3tan\theta}{2}$

Answer 2

$$\begin{lgathered}\sf \int \frac{d\theta}{4 {cos}^{2}\theta + 9{sin}^{2}\theta} \\ \\ \sf Dividing \: by \: {cos}^{2}\theta \\ \\ \sf \int \frac{{sec}^{2}\theta \: d\theta}{4 + 9{tan}^{2}\theta} \\ \\ \sf Substituting\: 3tan\theta = t \\ \\ \sf {sec}^{2}\theta \: d\theta = \frac{1}{3}dt \\ \\ \sf \int \frac{1}{3} \frac{dt}{4 + {t}^{2}} \\ \\ \sf \int \frac{1}{3}\frac {dt}{{2}^{2} + {t}^{2}} \\ \\ \sf \frac{1}{6} {tan}^{-1}\frac{t}{2} \\ \\ \sf \frac{1}{6} {tan}^{-1}\frac{3tan\theta}{2}\end{lgathered}$$