Question

Please solve this ASAP
and please no fake answers
​

Answer 1

$\rm\longrightarrow \displaystyle \large\int \rm \: \frac{sin \: x}{ {a}^{2} + {b}^{2} \: {cos}^{2} x } \: dx \\ \\ \rm let \: \: cos \: x = t \\ \rm - sin \: x \: dx =dt \\ \\ \\ \rm\longrightarrow \displaystyle \large\int \rm \: \dfrac{ - dt}{ {a}^{2} + {b}^{2} \: {t}^{2} } \: \\ \\ \\ \rm\longrightarrow \displaystyle \large - \dfrac{1}{{b}^{2}}\large\int \rm \: \dfrac{ dt}{ \dfrac{{a}^{2} }{{b}^{2}} + \: {t}^{2} } \: \\ \\ \\ \rm\longrightarrow \displaystyle \large - \rm \dfrac{1}{{b}^{2}}\rm \times \dfrac{b}{a} \times {tan}^{ - 1} \bigg(\dfrac{t \: b}{a} \bigg) + c\\ \\ \\ \rm\longrightarrow \displaystyle \large \rm \dfrac{ - 1}{{ab}} \times {tan}^{ - 1} \bigg(\dfrac{t \: b}{a} \bigg) + c\\ \\ \\ \rm \: now \: put \: \: t \: = cos \: x \\ \\ \\ \rm\longrightarrow \displaystyle \large \rm \dfrac{ - 1}{{ab}} \: \: \: {tan}^{ - 1} \bigg(\dfrac{ b \: cos \: x}{a} \bigg) + c$