Question

Solve this intrigation?

Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to evaluate the given integral.

$\displaystyle \rm = \int \dfrac{ \sin(2x) }{1 + \sin^{2} (x) } \: dx$

$\displaystyle \rm = \int \dfrac{2 \sin(x) \cos(x) }{1 + \sin^{2} (x) } \: dx$

Now let us assume that:

$\rm \longrightarrow u = 1 + { \sin}^{2}(x)$

$\rm \longrightarrow \dfrac{du}{dx} = 2 \sin(x) \cos(x)$

$\rm \longrightarrow dx = \dfrac{1}{2 \sin(x) \cos(x) } \: du$

Therefore, the integral becomes:

$\displaystyle \rm = \int \dfrac{1}{u} \: du$

$\displaystyle \rm = \ln(u) + C$

$\displaystyle \rm = \ln(1 + sin^{2}x ) + C$

Therefore:

$\displaystyle \rm \longrightarrow \int \dfrac{ \sin(2x) }{1 + \sin^{2} (x) } \: dx = \ln(1 + sin^{2}x ) + C$

★ Which is our required answer.

$\textsf{\large{\underline{Know More}:}}$

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$