Question

Consider the following statements
Statement I If $\int f(x) d x=g(x), $ then $\int f(x) g(x) d x$ 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 $\int f(x) g(x) d x$ 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, $\int f(x) d x=g(x)$
$\ \therefore \int f(x) g(x) d x $
$ =g(x) \int f(x) d x-\int\left[g^{\prime}(x) \int f(x) d x\right] d x $
[using integration by parts]
$ =g(x) g(x)-\int g^{\prime}(x) g(x) d x \left[\because \int f(x) d x=g(x)\right] $
$ =[g(x)]^2-\int f(x) g(x) d x$
$ 2 \int f(x) g(x) d x=[g(x)]^2$
$ \Rightarrow \int f(x) g(x) d x=\frac{1}{2}[g(x)]^2$
II. $\int f(x) g(x) d x=\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} e^{\sin ^{-1} x} d x $
Put, $ \sin ^{-1} x=t $
$\Rightarrow \frac{1}{\sqrt{1-x^2}} d x=d t $
$ \therefore \int f(x) g(x) d x=\int t e^t d t $
$=t \int e^t d t-\int\left[\frac{d}{d t}(t) \int e^t d t\right] d t$
$ =t e^t-\int e^t d t $
$ =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$