Q113 / 160 · AP EAMCET 2019 (20 Apr Shift 2) · Mathematics

$\int \frac{\sqrt{1-x^2} \sin ^{-1} x+x}{\sqrt{1-x^2}} d x=$

  • A
    $x \sin ^{-1} x+\sqrt{1-x^2}+c$
  • B
    $\sin ^{-1} x+\sqrt{1-x^2}+c$
  • Correct answer
    C
    $x \sin ^{-1} x+c$
  • D
    $\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+c$

Solution

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

Hence, the correct answer is option: 'C'.

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