Question

integrate (sin x)/(cos^2 x * sqrt(cos 2x)) dx​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\displaystyle\int\dfrac{sin(x)}{cos^2(x)\cdot\sqrt{cos(2x)}}\,dx}$

$\tt{\displaystyle=\int\dfrac{sin(x)}{cos^2(x)\cdot\sqrt{cos^2(x)-sin^2(x)}}\,dx}$

$\tt{\displaystyle=\int\dfrac{sin(x)}{cos^2(x)\cdot\,cos(x)\cdot\sqrt{1-\dfrac{sin^2(x)}{cos^2(x)}}}\,dx}$

$\tt{\displaystyle=\int\dfrac{sin(x)\cdot\,sec^2(x)}{cos(x)\cdot\sqrt{1-\dfrac{sin^2(x)}{cos^2(x)}}}\,dx}$

$\tt{\displaystyle=\int\dfrac{tan(x)\cdot\,sec^2(x)}{\sqrt{1-tan^2(x)}}\,dx}$

$\bf{Put\,\,\,1-tan^2(x)=t}$

$\bf{\implies\,-2\,tan(x)\,sex^2(x)\,dx=dt}$

$\bf{\implies\,tan(x)\,sex^2(x)\,dx=\dfrac{dt}{-2}}$

So,

$\tt{\displaystyle=\int\dfrac{dt}{(-2)\sqrt{t}}}$

$\tt{\displaystyle=-\dfrac{1}{2}\int\dfrac{dt}{\sqrt{t}}}$

$\tt{\displaystyle=-\dfrac{1}{2}\cdot2\sqrt{t}+C}$

$\tt{\displaystyle=-\sqrt{1-tan^2(x)}+C}$