Q.Integrate the function $\frac{\cos x}{\sqrt{4-\sin^2 x}}$

Concept: U Substitution — we substitute the inner function of the composition to simplify the square root.

Let $u = \sin x$, so $du = \cos x \, dx$. The integral becomes

This is a standard form: $\int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\left(\frac{u}{a}\right) + C$, with $a = 2$. …

Why U-Substitution Works Here

When you see a function like $\frac{\cos x}{\sqrt{4-\sin^2 x}}$, the key is to notice that the numerator $\cos x$ is the derivative of $\sin x$, which appears inside the square root. This is the classic signal for a u-substitution: if you set $u = \sin x$, then $du = \cos x \, dx$, and the entire integral simplifies into something you can handle with a standard formula.

The denominator $\sqrt{4 - \sin^2 x}$ becomes $\sqrt{4 - u^2}$, which is exactly the form that leads to an inverse sine (arcsine) integral. No messy trigonometric identities or integration by parts needed — just a clean substitution.

Step-by-Step Solution

1. Set up the substitution.

   Let $u = \sin x$. Then differentiate: $du = \cos x \, dx$. This directly replaces the numerator and the differential in the integral.

2. Rewrite the integral in terms of $u$.

   The original integral is:

Substituting $u = \sin x$ and $du = \cos x \, dx$ gives:

3. Recognize the standard form. The integral $\int \frac{du}{\sqrt{a^2 - u^2}}$ is a standard result. For $a > 0$, we have:

Here, $a^2 = 4$, so $a = 2$.

4. Apply the formula. With $a = 2$, the integral becomes:

5. Substitute back for $x$. Recall $u = \sin x$, so: $\sin^{-1}\left(\frac{\sin x}{2}\right) + C$ …

Method: $u$-substitution into the standard form $\int\frac{du}{\sqrt{a^2-u^2}}$

Use this when the numerator is (a constant times) the derivative of a function that appears squared under a $\sqrt{a^2-(\cdot)^2}$ radical — it maps directly onto an arcsine.

Steps

Step 1: Spot the derivative pair.

Identify $u$ so that $du$ matches the numerator. Here $\cos x\,dx=d(\sin x)$, so set $u=\sin x$.

Step 2: Rewrite in the standard radical form.

The integral becomes $\int\dfrac{du}{\sqrt{a^2-u^2}}$; read off $a$ from the constant under the root ($a^2=4\Rightarrow a=2$). …

Common Mistakes

Mistake 1: Forgetting the $a=2$ inside the arcsine.

Why it's wrong: $\int\dfrac{du}{\sqrt{4-u^2}}=\sin^{-1}\tfrac{u}{2}$, so the answer is $\sin^{-1}\tfrac{\sin x}{2}$, not $\sin^{-1}(\sin x)$. Correct approach: divide $u$ by $a=\sqrt4=2$.

Mistake 2: Misreading the sign under the root. …