Question

∫(1+sinx)/(1-sinx)dx ∫(1+sinx)/(1-sinx)dx @Mathematics

Answer 1

\[\frac{1+\sin(x)}{1-\sin(x)} \cdot \frac{1+\sin(x)}{1+\sin(x)}=\frac{(1+\sin(x))^2}{1-\sin^2(x)}=\frac{(1+\sin(x))^2}{\cos^2(x)}\]

Answer 2

\[\frac{1+2\sin(x)+\sin^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}+\frac{2\sin(x)}{\cos^2(x)}+\frac{\sin^2(x)}{\cos^2(x)}\]

Answer 3

\[\sec^2(x)+2 \cdot \frac{\sin(x)}{\cos^2(x)}+\tan^2(x)=\sec^2(x)+2\cdot \frac{\sin(x)}{\cos^2(x)}+(\sec^2(x)-1)\]

Answer 4

\[\int\limits_{}^{}\frac{1+\sin(x)}{1-\sin(x)} dx=\int\limits_{}^{}[\sec^2(x)+2 \cdot \frac{\sin(x)}{\cos^2(x)} +\sec^2(x)-1] dx\]

Answer 5

\[\tan(x)+\tan(x)-x+2\int\limits_{}^{}\frac{\sin(x)}{\cos^2(x) } dx\]

Answer 6

let u=cos(x) => du=-sin(x) dx => -du=sin(x) dx

Answer 7

\[2\tan(x)-x+2 \cdot \int\limits_{}^{}\frac{-du}{u^2}\]

Answer 8

\[2\tan(x)-x-2 \cdot \frac{u^{-2+1}}{-2+1}+C\]

Answer 9

\[2\tan(x)-x-2 \cdot \frac{u^{-1}}{-1}+C=2\tan(x)-x+\frac{2}{u}+C\]

Answer 10

replace u with cos(x) and you are done

Answer 11

okay thank you