Question

Assertion :Let $F\left(x\right)$ be an indefinite integral of $\frac{{\cos }^{3}x}{{\sin }^{2}x+\sin x}.x\ne n\pi$,
Statement 1 : The function $F\left(x\right)$ is $1-1$ on $(0,\pi /2].$ Reason: Statement 2 : $\log x$ increases from $-\infty$ to $0$ on $(0,1]$ and $\sin x$ increases from $0$ to $1$ on $(0,\pi /2]$.

• A. Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation for Statement 1
• B. Both Statement 1 and Statement 2 are correct but Statement 2 is not the correct explanation for Statement 1
• C. Statement 1 is correct but Statement 2 is incorrect
• D. Both Statement 1 and Statement 2 are incorrect

Answer 1

The correct option is A Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation for Statement 1

Given, $\frac{{\cos }^3\left(x\right)}{{\sin }^2\left(x\right)+\sin \left(x\right)}$
$=\frac{\cos \left(x\right)(1-{\sin }^2(x\left)\right)}{\sin \left(x\right)(1+\sin (x\left)\right)}$
$=\cot \left(x\right)(1-\sin (x\left)\right)$
$=\cot \left(x\right)-\cos \left(x\right)$
Hence
$\int \cot \left(x\right)-\cos \left(x\right).dx$
$F\left(x\right)=ln|\sin \left(x\right)|-\sin \left(x\right)+c$.
Now $\sin \left(x\right)$ is one-one on $(0,\frac{\pi }{2}]$ and also $0\le \sin \left(x\right)\le 1$ for $xϵ(0,\frac{\pi }{2})$.
Hence $ln|\sin \left(x\right)|$ increases from $-\infty$ to 0 for $xϵ(0,\frac{\pi }{2})$.
Therefore $ln|\sin \left(x\right)|$ is a one-one function for $xϵ(0,\frac{\pi }{2})$.
Thus, $F\left(x\right)$ is a one-one function in for $xϵ(0,\frac{\pi }{2})$.