Mathematics - Indefinite Integration Question with Solution | TestHub

$\int \frac{f(\mathrm{x})}{\cos \mathrm{x}}\mathrm{dx}=4{\mathrm{I}}_1-3{\mathrm{I}}_2$

${\mathrm{I}}_1=\int \frac{\cos x}{\sqrt{1-4{\sin }^{2}x}}dx=\frac{1}{2}{\sin }^{-1}(2\sin x)$

${\mathrm{I}}_2=\int \frac{\cos xdx}{\left(1-{\sin }^{2}x\right)\sqrt{1-4{\sin }^{2}x}}$

Put $\sin x=t$

$=\int \frac{dt}{\left(1-t^2\right)\sqrt{1-4t^2}}$

Put $t=\frac{1}{z}$

$\mathrm{dt}=-\frac{1}{{\mathrm{z}}^2}\mathrm{dz}$

Put $z^2=\alpha$

$$\begin{array}{ll}=-\int \frac{zdz}{\left(z^2-1\right)\sqrt{z^2-4}}& 2zdz=d\alpha \\ =-\frac{1}{2}\int \frac{d\alpha }{(\alpha -1)\sqrt{\alpha -4}}& \text{ Put }\alpha -4={\beta }^2\\ =-\int \frac{d\beta }{{\beta }^2+3}=-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{\beta }{\sqrt{3}}\right)=-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{\alpha -4}}{\sqrt{3}}\right)& \\ =\frac{-1}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{z^2-4}}{\sqrt{3}}\right)& \\ =-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{1-4t^2}}{\sqrt{3}t}\right)=-\frac{1}{3}{\tan }^{-1}\left(\frac{\sqrt{1-4{\sin }^{2}x}}{\sqrt{3}\sin x}\right)& \\ =\int \frac{f(x)}{\cos x}dx=2{\sin }^{-1}(2\sin x)+\sqrt{3}{\tan }^{-1}\left(\frac{f(x)}{\sqrt{3}\sin x}\right)& \end{array}$$