Question

$\int \frac{1}{ {sin}^{2}x + sin2x } dx$
give answer ..
​

Answer 1

$\huge \boxed{ \green{\mathfrak{solution}}}$

To find:

$\int \: \frac{1}{ {sin}^{2}x + sin2x } dx$

$\leadsto \int \frac{1}{ {sin}^{2}x + 2sinx \: cosx} dx$

Divide both numerator and denominator by cos^2x

$\leadsto \int \: \frac{ \frac{1}{ {cos}^{2} x} }{ \frac{ {sin}^{2} x}{ {cos}^{2} x} + \frac{2sinx \: cosx \: }{ {cos}^{2} x} } dx \\ \\ \leadsto \int \: \frac{ {sec}^{2}x }{ {tan}^{2}x + 2tanx }dx \\ \\ \ \:$

Now use substitution method ,

Let

◆tanx=t

=> sec^2dx =dt

$\leadsto \: \int \frac{dt}{ {t}^{2} + 2t } \\ \\ \leadsto \int \: \frac{dt}{t(t + 2)}$

Now we have use partial fraction method :

$\rightarrow\frac{1}{t(t + 2)} = \frac{a}{t} + \frac{b}{t + 2} \\ \\ \rightarrow \: a = \frac{1}{0 + 2} = \frac{1}{2} \\ and \: \\ \\ \rightarrow \: b = - \frac{1}{2}$

Now put the value of a and b

$\leadsto \frac{1}{2} \int ( \frac{1}{t} - \frac{1}{t + 2} )dt \\ \\ \leadsto \: \frac{1}{2} ( log(t )- log(t + 2) ) \\ \\ \leadsto \: \frac{1}{2} log( \frac{tanx}{tan(x + 2)} )$

Formula used :

$\boxed{ \bold{ \blue{ \int \frac{1}{x} dx = log(x) }}}$

$\boxed{ \blue{ \bold{ \frac{d}{dx} tanx = {sec}^{2 \: } x}}}$