Question

If $\int \csc 2xdx=f\left(g\right(x\left)\right)+c$ then

• A. range $g\left(x\right)=(-\infty ,\infty )$
• B. domain $f\left(x\right)=(-\infty ,\infty )$
• C. $g^{\prime }\left(x\right)={\sec }^{2}x$
• D. $f^{\prime }\left(x\right)=1/x$ for all $x\in (0,\infty )$

Answer 1

Answer: (A), (C), (D)

The correct options are
A range $g\left(x\right)=(-\infty ,\infty )$
C $g^{\prime }\left(x\right)={\sec }^{2}x$
D $f^{\prime }\left(x\right)=1/x$ for all $x\in (0,\infty )$

$\int \frac{1}{sin2x}.dx$

$=\int \frac{1}{2cosx.sinx}.dx$

Let

$sinx=t$

$cosxdx=dt$

$dx=\frac{dt}{cosx}$

Therefore

$I=\int \frac{1}{2sinx.cosx}.\frac{dt}{cosx}$

$=\int \frac{1}{2sinxco{s}^{2}x}.dt$

$=\int \frac{1}{2t(1-t^2)}.dt$

$=\int \frac{1-t^2+t^2}{2t(1-t^2)}.dt$

$=\int \frac{1}{2t}+\frac{t}{2(1-t^2)}.dt$

$=\frac{lnt}{2}+\frac{1}{4}\int \frac{2t}{1-t^2}.dt$

$=\frac{lnt}{2}-\frac{ln(1-t^2)}{4}+C$

$=\frac{ln\left(t^2\right)-ln(1-t^2)}{4}+C$

$=\frac{ln\left(\frac{t^2}{1-t^2}\right)}{4}+C$

$=\frac{ln\left(\frac{si{n}^{2}x}{co{s}^{2}x}\right)}{4}+C$

$=\frac{2ln\left(tanx\right)}{4}+C$

$=\frac{ln\left(tanx\right)}{2}+C$

Hence $f\left(x\right)=lnx$ and $g\left(x\right)=tanx$

$f^{\prime }\left(x\right)=\frac{1}{x}$ and $g^{\prime }\left(x\right)=se{c}^{2}x$.

Also the domain and range of $g\left(x\right)$ is $(-\infty ,\infty )$