1708\hspace{-20}\bf{(2):\; }$Let $\bf{\mathbb{I}=\int\frac{\sin x}{\sqrt{1-\sin x}}dx = \int\frac{\sin x}{\sqrt{1-\sin x}}\times \frac{\sqrt{1+\sin x}}{\sqrt{1+\sin x}}dx}$\\\\\\ So Integral $\bf{\mathbb{I}=\int\frac{\sin x}{\cos x}\cdot \sqrt{1+\sin x}\; dx}$\\\\\\ Now Let $\bf{(1+\sin x)=t^2\;,}$ Then $\bf{\cos xdx = 2tdt\Rightarrow dx = \frac{2t}{\cos x}dt}$\\\\\\ So Integral $\bf{\mathbb{I}=\int\frac{\sin x}{\cos x}\cdot \frac{2t^2}{\cos x}dt = \int\frac{\sin x}{1-\sin^2 x}\cdot 2t^2dt}$\\\\\\ So Integral $\bf{\mathbhbb{I}=\int\frac{(t^2-1)\cdot 2t^2}{1-\left(t^2-1\right)}dt=2\int\frac{(t^2-1)}{2-t^2}dt=-2\int\frac{(t^2-1)}{(t^2-2)}dt}$\\\\\\ So $\bf{I = -2\left\{\int\frac{(t^2-2)+1}{(t^2-2)}\right\}dt = -2\int 1\cdo dt-2\int\frac{1}{t^2-\left(\sqrt{2}\right)^2}dt}$\\\\\\ So $\bf{I = -2t-2\cdot \frac{1}{2\sqrt{2}}\ln\left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|+\mathbb{C}.....}$\\\\\\
\hspace{-20}$So $\bf{\mathbb{I}=\int\frac{\sin x}{\sqrt{1-\sin x}}dx=-2\sqrt{1+\sin x}-\frac{1}{\sqrt{2}}\cdot \ln\left|\frac{\sqrt{1+\sin x}-\sqrt{2}}{\sqrt{1+\sin x}+\sqrt{2}}\right|+\mathbb{C}....}$
1708
man111 singh
·2014-06-27 18:09:31
\hspace{-20}\bf{(1):\; }$Let $\bf{\mathbb{I}=\int\frac{(\cos x-\sin x)}{\sqrt{\sin 2x}}dx = \int\frac{(\cos x-\sin x)}{\sqrt{\left(\cos x+\sin x\right)^2-1}}dx}$\\\\\\ Now Let $\bf{(\cos x+\sin x)=t\;,}$ Then $\bf{(\cos x-\sin x)dx = dt}$\\\\\\ So Integral Convert into $\bf{\mathbb{I}=\int\frac{1}{\sqrt{t^2-1}}dt=\ln \left|t+\sqrt{t^2-1}\right|+\mathbb{C}}$\\\\\\ So $\bf{\mathbb{I}=\int\frac{(\cos x-\sin x)}{\sqrt{\sin 2x}}dx=\ln \left|\left(\cos x+\sin x\right)+\sqrt{\left(\cos x+\sin x\right)^2-1}\right|+\mathbb{C}}$\\\\\\ So $\bf{\mathbb{I}=\int\frac{(\cos x-\sin x)}{\sqrt{\sin 2x}}dx=\ln \left|\left(\cos x+\sin x\right)+\sqrt{\sin 2x}\right|+\mathbb{C}}$
