Q.Evaluate: $\int_{0}^{1} \dfrac{dx}{(1+x^2)\sqrt{1-x^2}}$ (Hint: let $x=\sin\theta$)

Concept: U Substitution (trigonometric substitution). The hint $x = \sin\theta$ simplifies the square root and the limits.

Let $x = \sin\theta$, so $dx = \cos\theta \, d\theta$. When $x=0$, $\theta=0$; when $x=1$, $\theta=\frac{\pi}{2}$.

The integral becomes:

Divide numerator and denominator by $\cos^2\theta$:

$\int_{0}^{\pi/2} \frac{\sec^2\theta \, d\theta}{\sec^2\theta + \tan^2\theta} = \int_{0}^{\pi/2} \frac{\sec^2\theta \, d\theta}{1 + 2\tan^2\theta}.$ …

Why this substitution works

When you see a $\sqrt{1-x^2}$ in the denominator, your first instinct should be a trigonometric substitution. The hint suggests $x = \sin\theta$, which is perfect because:

• $\sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \cos\theta$ (for $\theta$ in $[0, \pi/2]$, where cosine is non-negative)
• $dx = \cos\theta\,d\theta$
• The limits $x=0$ and $x=1$ become $\theta=0$ and $\theta=\pi/2$

The $1+x^2$ in the denominator becomes $1+\sin^2\theta$, which is a clean expression. This turns a messy-looking integral into something we can handle with standard techniques.

Step-by-step solution

1. Apply the substitution

Let $x = \sin\theta$, so $dx = \cos\theta\,d\theta$. When $x=0$, $\theta=0$; when $x=1$, $\theta=\pi/2$.

The integral becomes:

2. Simplify the integrand

The $\cos\theta$ cancels (provided $\cos\theta \neq 0$, which is true except at the endpoint $\theta=\pi/2$ — a removable issue):

3. Transform $\sin^2\theta$ into a form we can integrate

We know $\sin^2\theta = 1 - \cos^2\theta$, but that doesn't help directly. Instead, use the identity $\sin^2\theta = \frac{1-\cos 2\theta}{2}$:

So the integral is:

4. Substitute again to handle the $\cos 2\theta$ term

Let $t = 2\theta$, so $d\theta = dt/2$. When $\theta=0$, $t=0$; when $\theta=\pi/2$, $t=\pi$.

5. Use the tangent half-angle substitution

This is the classic method for integrals of the form $\int \frac{dt}{a + b\cos t}$. Let $u = \tan(t/2)$. Then:

• $\cos t = \frac{1-u^2}{1+u^2}$
• $dt = \frac{2\,du}{1+u^2}$
• When $t=0$, $u=0$; when $t=\pi$, $u \to \infty$

Method: Trigonometric substitution $x=\sin\theta$

Use this for definite integrals containing $\sqrt{1-x^2}$, especially with an extra $1+x^2$ factor. Setting $x=\sin\theta$ clears the root and turns the integral into a manageable trig one.

Steps

Step 1: Substitute $x=\sin\theta$.

Step 2: Rewrite the integrand.

Step 3: Reduce $1+\sin^2\theta$ using a further substitution.

Divide by $\cos^2\theta$ and let $t=\tan\theta$: $\int\dfrac{\sec^2\theta\,d\theta}{\sec^2\theta+\tan^2\theta}=\int\dfrac{dt}{1+2t^2}$. …

Common Mistakes

Mistake 1: Not choosing $x=\sin\theta$.

Why it's wrong: $\sqrt{1-x^2}$ blocks algebraic substitution; $x=\sin\theta$ turns it into $\cos\theta$ and cancels cleanly. Correct approach: use the trig substitution first.

Mistake 2: Getting stuck at $\int\frac{d\theta}{1+\sin^2\theta}$.

Why it's wrong: this needs a second step — divide by $\cos^2\theta$ and set $t=\tan\theta$ to reach $\int\tfrac{dt}{1+2t^2}$. Correct approach: convert to $\tan\theta$ form.

Mistake 3: Forgetting the $\tfrac{1}{\sqrt2}$ factor / limits. …