Question

int sin^{-1}x dx

Explanation:

Step1: Use Integration by Parts

Recall the integration by parts formula: $\int u \, dv = uv - \int v \, du$. Let $u = \sin^{-1}x$ and $dv = dx$. Then we need to find $du$ and $v$.

• For $u = \sin^{-1}x$, the derivative $du$ is $\frac{1}{\sqrt{1 - x^2}}dx$ (by the derivative formula for the inverse sine function).
• For $dv = dx$, integrating both sides gives $v = x$ (since $\int dx = x + C$, and we can ignore the constant of integration for now as we will add it at the end).

Step2: Apply the Integration by Parts Formula

Substitute $u$, $dv$, $du$, and $v$ into the integration by parts formula:
$\int \sin^{-1}x \, dx = uv - \int v \, du = x\sin^{-1}x - \int x \cdot \frac{1}{\sqrt{1 - x^2}}dx$

Step3: Integrate $\int \frac{x}{\sqrt{1 - x^2}}dx$

Let $t = 1 - x^2$. Then the derivative $dt = -2x \, dx$, which implies $-\frac{1}{2}dt = x \, dx$.
Substitute $t$ and $dt$ into the integral:
$\int \frac{x}{\sqrt{1 - x^2}}dx = \int \frac{1}{\sqrt{t}} \cdot (-\frac{1}{2})dt = -\frac{1}{2} \int t^{-\frac{1}{2}}dt$
Integrate $t^{-\frac{1}{2}}$: $\int t^{-\frac{1}{2}}dt = 2t^{\frac{1}{2}} + C = 2\sqrt{t} + C$
Substitute back $t = 1 - x^2$:
$-\frac{1}{2} \cdot 2\sqrt{1 - x^2} + C = -\sqrt{1 - x^2} + C$

Step4: Combine the Results

Substitute the result of the integral from Step 3 back into the expression from Step 2:
$\int \sin^{-1}x \, dx = x\sin^{-1}x - (-\sqrt{1 - x^2}) + C = x\sin^{-1}x + \sqrt{1 - x^2} + C$

Answer:

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