Evaluate ∫ sin^-1 x dx.

Question 89

HardOriginal question

Evaluate $\int \sin^{-1} x\,dx$ .

x\sin^{-1} x - \sqrt{1 - x^{2}} + C

Correct answerYour answer

\frac{1}{\sqrt{1 - x^{2}}} + C

Correct answerYour answer

x\sin^{-1} x + \sqrt{1 - x^{2}} + C

Correct answerYour answer

x\sin^{-1} x + C

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

Write it as $\int 1 \cdot \sin^{-1} x\,dx$ with $u = \sin^{-1} x$ , $dv = dx$ : $x\sin^{-1} x - \int \frac{x}{\sqrt{1 - x^{2}}}\,dx$ .
$\int \frac{x}{\sqrt{1 - x^{2}}}\,dx = -\sqrt{1 - x^{2}}$ , so the answer is $x\sin^{-1} x + \sqrt{1 - x^{2}} + C$ .

Worth noting

$-\sqrt{1 - x^{2}}$ loses the sign: subtracting $-\sqrt{1 - x^{2}}$ adds it.

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