Question

Consider the following statements
Statement I If $∫f(x)dx=g(x),$ then $∫f(x)g(x)dx$ is equal to $\frac{1}{2}[g(x)]$.
Statement II If $f(x)=\frac{{\sin }^{−1}x}{\sqrt{1−x^2}}$ and $g(x)=e^{{\sin }^{−1}x}$, then $∫f(x)g(x)dx$ is equal to $e^{{\sin }^{−1}x}\left({\sin }^{−1}x+1\right)+C$
Choose the correct option.

• A. Statement I is true
• B. Statement II is true
• C. Both statements are true
• D. Both statements are false

Answer 1

Answer: (D)

I. Given, $∫f(x)dx=g(x)$
$∴∫f(x)g(x)dx$
$=g(x)∫f(x)dx−∫\left[g^{′}(x)∫f(x)dx\right]dx$
[using integration by parts]
$=g(x)g(x)−∫g^{′}(x)g(x)dx\left[âˆ\mu ∫f(x)dx=g(x)\right]$
$=[g(x){]}^2−∫f(x)g(x)dx$
$2∫f(x)g(x)dx=[g(x){]}^2$
$⇒∫f(x)g(x)dx=\frac{1}{2}[g(x){]}^2$
II. $∫f(x)g(x)dx=∫\frac{{\sin }^{−1}x}{\sqrt{1−x^2}}{e}^{{\sin }^{−1}x}dx$
Put, ${\sin }^{−1}x=t$
$⇒\frac{1}{\sqrt{1−x^2}}dx=dt$
$∴∫f(x)g(x)dx=∫t{e}^{t}dt$
$=t∫e^{t}dt−∫\left[\frac{d}{dt}(t)∫e^{t}dt\right]dt$
$=t{e}^t−∫e^{t}dt$
$=t{e}^t−e^t+C$
$=e^{{\sin }^{−1}x}{\sin }^{−1}x−e^{{\sin }^{−1}x}+C$
$=e^{{\sin }^{−1}x}\left({\sin }^{−1}x−1\right)+C$