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$\int \frac{\sqrt{1-x^2} \sin ^{-1} x+x}{\sqrt{1-x^2}} d x=$
Options:
Solution:
1554 Upvotes
Verified Answer
The correct answer is:
$x \sin ^{-1} x+c$
$$
\begin{aligned}
& \text { Given, } \int \frac{\sqrt{1-x^2} \cdot \sin ^{-1} x+x}{\sqrt{1-x^2}} d x \\
& \int\left(\frac{\sqrt{1-x^2} \cdot \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}\right) d x \\
& \int\left(\sin ^{-1} x+\frac{x}{\sqrt{1-x^2}}\right) d x \\
& \int \sin ^{-1} x d x+\int \frac{x}{\sqrt{1-x^2}} d x \\
& \sin ^{-1} x \cdot \int 1 d x-\int\left(\frac{d}{d x} \sin ^{-1} x \cdot \int 1 d x\right) d x+\int \frac{x}{\sqrt{1-x^2}} d x \\
& \quad\left[\because \text { By using by parts with } f(x)=\sin ^{-1} x ; g(x)=1\right] \\
& \sin ^{-1} x \cdot x-\int \frac{1}{\sqrt{1-x^2}} \cdot x d x+\int \frac{x}{\sqrt{1-x^2}} d x+c \\
& x \cdot \sin ^{-1} x+c
\end{aligned}
$$
$\therefore$ Hence option (c) is correct.
\begin{aligned}
& \text { Given, } \int \frac{\sqrt{1-x^2} \cdot \sin ^{-1} x+x}{\sqrt{1-x^2}} d x \\
& \int\left(\frac{\sqrt{1-x^2} \cdot \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}\right) d x \\
& \int\left(\sin ^{-1} x+\frac{x}{\sqrt{1-x^2}}\right) d x \\
& \int \sin ^{-1} x d x+\int \frac{x}{\sqrt{1-x^2}} d x \\
& \sin ^{-1} x \cdot \int 1 d x-\int\left(\frac{d}{d x} \sin ^{-1} x \cdot \int 1 d x\right) d x+\int \frac{x}{\sqrt{1-x^2}} d x \\
& \quad\left[\because \text { By using by parts with } f(x)=\sin ^{-1} x ; g(x)=1\right] \\
& \sin ^{-1} x \cdot x-\int \frac{1}{\sqrt{1-x^2}} \cdot x d x+\int \frac{x}{\sqrt{1-x^2}} d x+c \\
& x \cdot \sin ^{-1} x+c
\end{aligned}
$$
$\therefore$ Hence option (c) is correct.
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