Question

Integrate this Integrand wrt x​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{I=\displaystyle\int\dfrac{2\,sin(x)}{(3+sin(2x))}\,dx}$

$\tt{\implies\,I=\displaystyle\int\dfrac{2\,sin(x)}{3+sin(2x)}\,dx}$

$\tt{\implies\,I=\displaystyle\int\dfrac{2\,sin(x)}{3+sin(2x)}\,dx}$

$\tt{\implies\,I=\displaystyle\int\dfrac{sin(x)-cos(x)+sin(x)+cos(x)}{3+sin(2x)}\,dx}$

$\tt{\implies\,I=\displaystyle\int\dfrac{sin(x)+cos(x)}{3+sin(2x)}\,dx-\int\dfrac{cos(x)-sin(x)}{3+sin(2x)}\,dx}$

$\tt{\implies\,I=I_{1}-I_{2}}$

$\bold{\sf{\bullet\,\,\green{Solving\,\,\,\tt{I_{1}}}\,:}}$

$\tt{\implies\,I_{1}=\displaystyle\int\dfrac{sin(x)+cos(x)}{3+sin(2x)}\,dx}$

$\tt{\implies\,I_{1}=\displaystyle\int\dfrac{sin(x)+cos(x)}{4-1+sin(2x)}\,dx}$

$\tt{\implies\,I_{1}=\displaystyle\int\dfrac{sin(x)+cos(x)}{4-1+2\,sin(x)\,cos(x)}\,dx}$

$\tt{\implies\,I_{1}=\displaystyle\int\dfrac{sin(x)+cos(x)}{4-\{1-2\,sin(x)\,cos(x)\}}\,dx}$

$\tt{\implies\,I_{1}=\displaystyle\int\dfrac{sin(x)+cos(x)}{4-\{sin(x)-cos(x)\}^2}\,dx}$

$\sf{\blue{Put\,\,\, sin(x)-cos(x)=t}}$

$\sf{\implies\blue{ (cos(x)+sin(x))dx=dt}}$

So,

$\tt{I_{1}=\displaystyle\int\dfrac{dt}{4-t^2}}$

$\tt{\implies\,I_{1}=\displaystyle\int\dfrac{dt}{(2)-(t)^2}}$

$\tt{\implies\,I_{1}=\ln\left|\dfrac{2+t}{2-t}\right|+c_{1}}$

$\tt{\implies\,I_{1}=\ln\left|\dfrac{2+sin(x)-cos(x)}{2-sin(x)+cos(x)}\right|+c_{1}}$

$\bold{\sf{\bullet\,\,\green{Solving\,\,\,\tt{I_{2}}}\,:}}$

$\tt{I_{2}=\displaystyle\int\dfrac{cos(x)-sin(x)}{3+sin(2x)}\,dx}$

$\tt{\implies\,I_{2}=\displaystyle\int\dfrac{cos(x)-sin(x)}{2+1+2\,sin(x)\,cos(x)}\,dx}$

$\tt{\implies\,I_{2}=\displaystyle\int\dfrac{cos(x)-sin(x)}{2+\{sin(x)+cos(x)\}^{2}}\,dx}$

$\sf{\blue{Put\,\,\, sin(x)+cos(x)=u}}$

$\sf{\implies\blue{(cos(x)-sin(x))\,dx=du}}$

So,

$\tt{I_{2}=\displaystyle\int\dfrac{du}{2+(u)^{2}}}$

$\tt{\implies\,I_{2}=\displaystyle\int\dfrac{du}{\left(\sqrt{2}\right)^2+\left(u\right)^{2}}}$

$\tt{\implies\,I_{2}=\dfrac{1}{\sqrt{2}}\,tan^{-1}\left(\dfrac{u}{\sqrt{2}}\right)}$

$\tt{\implies\,I_{2}=\dfrac{1}{\sqrt{2}}\,tan^{-1}\left\{\dfrac{sin(x)+cos(x)}{\sqrt{2}}\right\}+c_{2}}$

Now,

$\tt{\implies\,I=I_{1}-I_{2}}$

$\tt{\implies\,I=\ln\left|\dfrac{2+sin(x)-cos(x)}{2-sin(x)+cos(x)}\right|+c_{1}-\dfrac{1}{\sqrt{2}}\,tan^{-1}\left\{\dfrac{sin(x)+cos(x)}{\sqrt{2}}\right\}-c_{2}}$

Put $\sf{C=c_{1}-c_{2}}$

$\tt{\implies\,I=\ln\left|\dfrac{2+sin(x)-cos(x)}{2-sin(x)+cos(x)}\right|-\dfrac{1}{\sqrt{2}}\,tan^{-1}\left\{\dfrac{sin(x)+cos(x)}{\sqrt{2}}\right\}+C}$