Q.Prove that $\int_{0}^{1}\sin^{-1}x\,dx=\frac{\pi}{2}-1$

Concept: Integration by parts — we treat $\sin^{-1}x$ as a function whose derivative is simpler, and integrate $1$ as the second function.

Let $I = \int_{0}^{1} \sin^{-1}x \, dx$. Write it as $\int_{0}^{1} \sin^{-1}x \cdot 1 \, dx$.

Using integration by parts:

$\int u\,dv = uv - \int v\,du$, with $u = \sin^{-1}x$, $dv = dx$.

Then $du = \frac{1}{\sqrt{1-x^2}}\,dx$, $v = x$.

So

$I = \left[ x \sin^{-1}x \right]_{0}^{1} - \int_{0}^{1} \frac{x}{\sqrt{1-x^2}}\,dx$.

Evaluate the boundary term: at $x=1$, $\sin^{-1}1 = \frac{\pi}{2}$, so $1 \cdot \frac{\pi}{2} = \frac{\pi}{2}$; at $x=0$, $0 \cdot 0 = 0$. Hence the first part is $\frac{\pi}{2}$. …

The problem asks us to prove a definite integral involving the inverse sine function. At first glance, you might wonder: what’s the antiderivative of $\sin^{-1}x$? It’s not a standard derivative we memorize. So we need a technique to break it down.

The key insight is integration by parts. This method is perfect when we have a product of two functions — and here, we can think of $\sin^{-1}x$ as $1 \cdot \sin^{-1}x$. The “1” is easy to integrate, and the derivative of $\sin^{-1}x$ is a rational function, which simplifies things.

Let’s work through it step by step.

1. Choose $u$ and $dv$.

   We set $u = \sin^{-1}x$ and $dv = 1 \, dx$.

   Why? Because $du$ will become $\frac{1}{\sqrt{1-x^2}} dx$, a simpler expression, and $v$ will just be $x$.

2. Compute $du$ and $v$.

   • $du = \frac{1}{\sqrt{1-x^2}} \, dx$
   • $v = \int 1 \, dx = x$

3. Apply integration by parts.

4. Evaluate the boundary term.

   At $x=1$: $1 \cdot \sin^{-1}(1) = 1 \cdot \frac{\pi}{2} = \frac{\pi}{2}$

   At $x=0$: $0 \cdot \sin^{-1}(0) = 0$

   So the boundary term is $\frac{\pi}{2} - 0 = \frac{\pi}{2}$.

5. Simplify the remaining integral.

   We now need $\int_{0}^{1} \frac{x}{\sqrt{1-x^2}} \, dx$.

   Notice the numerator $x$ is almost the derivative of $1-x^2$ (which is $-2x$). This suggests a substitution.

   Let $t = 1 - x^2$. Then $dt = -2x \, dx$, so $x \, dx = -\frac{1}{2} dt$.

   When $x=0$, $t=1$; when $x=1$, $t=0$.

   The integral becomes:

Flipping the limits removes the minus sign:

6. Integrate with respect to $t$. …

Method: Integration by parts on a lone inverse-trig function

Use this for $\int\sin^{-1}x\,dx$ (or $\cos^{-1}x,\tan^{-1}x$): write it as $1\cdot\sin^{-1}x$ so parts can differentiate the inverse trig into a rational function.

Steps

Step 1: Set up parts.

Take $u=\sin^{-1}x$, $dv=dx$, so $du=\dfrac{dx}{\sqrt{1-x^2}}$ and $v=x$.

Step 2: Apply the formula. …

Common Mistakes

Mistake 1: Not writing $\sin^{-1}x$ as $1\cdot\sin^{-1}x$.

Why it's wrong: without a $dv=dx$ factor, integration by parts cannot start on a lone inverse-trig. Correct approach: take $u=\sin^{-1}x$, $dv=dx$.

Mistake 2: Wrong derivative of $\sin^{-1}x$.

Why it's wrong: $\tfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}$; an error here corrupts the leftover integral. Correct approach: use the standard derivative. …