Q.Integrate the following function: $\sin^{-1} (\cos x)$

The key idea is to simplify $\sin^{-1}(\cos x)$ using the relationship between inverse trigonometric functions and complementary angles.

Step 1: Recall the identity $\cos x = \sin\left(\frac{\pi}{2} - x\right)$.

Step 2: Substitute into the function:

$\sin^{-1}(\cos x) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - x\right)\right)$ …

The Core Idea: Why This Isn't Just Cancellation

When you see $\sin^{-1}(\sin \theta)$, the natural instinct is to say it equals $\theta$. But that's only true if $\theta$ is in the principal value range of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Our integrand is $\sin^{-1}(\cos x)$, not $\sin^{-1}(\sin x)$. So the first job is to rewrite $\cos x$ as a sine of something, and then carefully check whether that "something" falls inside $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

The identity $\cos x = \sin(\frac{\pi}{2} - x)$ is the natural bridge. But $\frac{\pi}{2} - x$ can be any real number, not necessarily in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. So we must handle the domain.

Step-by-Step Solution

1. Rewrite the integrand using a cofunction identity.

We know:

Therefore:

2. Understand when $\sin^{-1}(\sin y) = y$ holds.

The function $\sin^{-1}$ returns an angle in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. So:

If $y$ is outside this interval, $\sin^{-1}(\sin y)$ gives the equivalent angle inside $[-\frac{\pi}{2}, \frac{\pi}{2}]$ that has the same sine.

3. Determine the range of $y = \frac{\pi}{2} - x$.

Since $x$ is a real variable (the integration variable), $y = \frac{\pi}{2} - x$ can be any real number. So we cannot simply say $\sin^{-1}(\sin y) = y$ for all $x$.

However, in indefinite integration, we typically assume $x$ is such that the expression is well-defined and we can find an antiderivative valid on an interval. The standard approach is to work on an interval where $y$ lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

4. Find the interval for $x$ where $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

We need:

Subtract $\frac{\pi}{2}$ from all parts:

Multiply by $-1$ (reversing inequalities):

So on the interval $x \in [0, \pi]$, we have $\frac{\pi}{2} - x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, and therefore: …

Method: Convert an inverse-trig-of-trig into a linear expression

An expression like $\sin^{-1}(\cos x)$ is not integrated as an inverse function; use a co-function identity to rewrite it as a simple linear expression in $x$ (valid on the stated interval), then integrate directly.

Steps

Step 1: Rewrite the inside using a co-function shift.

Step 2: Cancel $\sin^{-1}$ against $\sin$ on the valid range.

$\sin^{-1}(\cos x)=\sin^{-1}\!\left(\sin\!\left(\tfrac{\pi}{2}-x\right)\right)=\tfrac{\pi}{2}-x$ …

Common Mistakes

Mistake 1: Treating $\sin^{-1}(\cos x)$ as needing integration by parts as an inverse function.

Why it's wrong: it first simplifies to a linear expression, making parts unnecessary. Correct approach: use $\cos x=\sin(\tfrac{\pi}{2}-x)$ so $\sin^{-1}(\cos x)=\tfrac{\pi}{2}-x$ on the valid interval.

Mistake 2: Ignoring the principal-value range where the cancellation holds. …