Q.Integrate the following function: $\sqrt{1-4x^2}$

1. Simplify the denominator. Using $\cos^2 x=1-\sin^2 x$,

2. Substitute $t=\sin x$, so $dt=\cos x\,dx$:

3. Reduce to standard form. Since $25-16t^2=25\!\left(1-\big(\tfrac{4t}{5}\big)^2\right)$,

Let $u=\tfrac{4t}{5}$, $dt=\tfrac54\,du$:

4. Back-substitute $t=\sin x$. A minus sign under the root means an inverse sine, not an inverse hyperbolic sine.

Check: $\dfrac{d}{dx}\Big[\tfrac14\sin^{-1}\!\big(\tfrac{4\sin x}{5}\big)\Big]=\dfrac{\cos x}{\sqrt{25-16\sin^2 x}}=\dfrac{\cos x}{\sqrt{16\cos^2 x+9}}$, matching the integrand.

The integral $\int \sqrt{x^2 + 2x} \, dx$ involves the square root of a quadratic expression. Integrals of the form $\int \sqrt{ax^2 + bx + c} \, dx$ are typically solved by completing the square for the quadratic expression $ax^2 + bx + c$. This transforms the integral into one of three standard forms: $\int \sqrt{a^2 - u^2} \, du$, $\int \sqrt{u^2 + a^2} \, du$, or $\int \sqrt{u^2 - a^2} \, du$, for which direct formulas exist.

In this problem, we will complete the square for $x^2 + 2x$ and then apply the appropriate standard integral formula.

1. Complete the square for the expression under the square root: The quadratic expression is $x^2 + 2x$. To complete the square, we add and subtract the square of half the coefficient of $x$. The coefficient of $x$ is $2$, so half of it is $1$, and its square is $1^2 = 1$.

Now the integral becomes $\int \sqrt{(x+1)^2 - 1^2} \, dx$.

2. Identify the standard integral form:

Let $u = x+1$. Then $du = dx$. The integral transforms to $\int \sqrt{u^2 - 1^2} \, du$.

This is of the form $\int \sqrt{u^2 - a^2} \, du$, where $a=1$.

3. Apply the standard integral formula: The standard formula for $\int \sqrt{y^2 - a^2} \, dy$ is:

   $\int \sqrt{y^2 - a^2} \, dy = \frac{y}{2}\sqrt{y^2 - a^2} - \frac{a^2}{2}\cosh^{-1}\left(\frac{y}{a}\right) + C$

   Applying this formula with $y=u$ and $a=1$:

> [!WARNING]
> Be careful with the sign and the inverse hyperbolic function. For $\sqrt{y^2 - a^2}$, it's $\cosh^{-1}$ with a negative sign. For $\sqrt{y^2 + a^2}$, it's $\sinh^{-1}$ with a positive sign.

4. Substitute back and simplify:

Substitute $u = x+1$ back into the expression:

Simplify the term under the square root: $(x+1)^2 - 1 = x^2 + 2x + 1 - 1 = x^2 + 2x$.
So the integral evaluates to:

5. Compare with the given options:

   Comparing our result with the provided options:

   (A) $\dfrac{x+1}{2} \sqrt{x^2 + 2x} + \dfrac{1}{2} \sinh^{-1} \left( \dfrac{x+1}{2} \right) + C$

   (B) $\dfrac{x+1}{2} \sqrt{x^2 + 2x} + \dfrac{1}{2} \sinh^{-1} (x+1) + C$

   (C) $\dfrac{x+1}{2} \sqrt{x^2 + 2x} - \dfrac{1}{2} \cosh^{-1} \left( \dfrac{x+1}{2} \right) + C$

   (D) $\dfrac{x+1}{2} \sqrt{x^2 + 2x} - \dfrac{1}{2} \cosh^{-1} (x+1) + C$

   Our derived expression matches option (D). The condition $x \geq 0$ ensures that $x+1 \geq 1$, so $\cosh^{-1}(x+1)$ is well-defined.

The standard form of this integral splits into an inverse-hyperbolic-secant part and an arcsine part. Writing the integrand as

and integrating each term:

since $\dfrac{d}{dx}\operatorname{Sech}^{-1}\sqrt{x}=-\dfrac{1}{2x\sqrt{1-x}}$ and $\dfrac{d}{dx}\sin^{-1}\sqrt{x}=\dfrac{1}{2\sqrt{x}\sqrt{1-x}}$.

Comparing with the given form $2f(x)-2\sin^{-1}\sqrt{x}+c$, the non-arcsine part is the inverse-hyperbolic-secant term, so

(Note: as printed, the radical in the integrand appears mistyped; the arcsine term in the stated result requires a $\sqrt{1-x}$ factor. The result above matches the intended standard integral and the exam key.)

Concept. A half-angle trigonometric substitution rationalises $\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}$, because $\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}} = \tan\dfrac{\theta}{2}$.

Step 1 — substitute. Let $\sqrt{x}=\cos\theta$, $\theta\in(0,\tfrac{\pi}{2})$, so $x=\cos^2\theta$, $dx=-2\cos\theta\sin\theta\,d\theta$:

Step 2 — simplify with half-angles. $\tan\dfrac{\theta}{2}\,\sin\theta = 2\sin^2\dfrac{\theta}{2} = 1-\cos\theta$, hence

Step 3 — return to $x$. With $\theta=\cos^{-1}\sqrt{x}$: $2\theta = \pi - 2\sin^{-1}\sqrt{x}$ (the $\pi$ is absorbed into the constant), $\sec\theta+\tan\theta = \dfrac{1+\sqrt{1-x}}{\sqrt{x}}$. So

Step 4 — identify $f$. Comparing with $I = 2f(x)-2\sin^{-1}\sqrt{x}+c$:

(using $(1+\sqrt{1-x})(1-\sqrt{1-x})=x$). This logarithm is precisely the inverse hyperbolic secant evaluated at $\sqrt{x}$ — the $\operatorname{sech}^{-1}\sqrt{x}$ form listed (the two log-branches of $\operatorname{sech}^{-1}y$ are $\log\frac{1\pm\sqrt{1-y^2}}{y}$). The $\csc^{-1}\sqrt{x}$ option is impossible on the domain $0<x<1$ (it needs $\sqrt{x}\ge 1$), and the plain log options differentiate to $\frac{1}{2x\sqrt{1+x}}$-type expressions, not the required $\frac{1}{2x\sqrt{1-x}}$.

ANSWER: D

The key here is that the denominator has a power of $3/2$, which is an awkward exponent. Direct substitution or partial fractions won't help. But if we can rewrite the expression so that the square root in the denominator becomes a simple trigonometric function, the integration becomes straightforward.

The form $x^2 - a^2$ under a square root (or a power of it) is a classic signal for the substitution $x = a \sec \theta$. Why? Because $\sec^2 \theta - 1 = \tan^2 \theta$, so $x^2 - a^2 = a^2 \tan^2 \theta$, and the square root becomes $a |\tan \theta|$. For $x > a$ (the usual domain), $\tan \theta > 0$, so we can drop the absolute value.

Let's work through it.

1. Substitute $x = a \sec \theta$.

   Then $dx = a \sec \theta \tan \theta \, d\theta$.

   Also, $x^2 - a^2 = a^2 (\sec^2 \theta - 1) = a^2 \tan^2 \theta$.

   Therefore $(x^2 - a^2)^{3/2} = (a^2 \tan^2 \theta)^{3/2} = a^3 |\tan \theta|^3$. For $\theta \in (0, \pi/2)$ (so $x > a$), $\tan \theta > 0$, so this is $a^3 \tan^3 \theta$.

2. Rewrite the integral.

3. Simplify the trigonometric expression. $\frac{\sec \theta}{\tan^2 \theta} = \frac{1/\cos \theta}{\sin^2 \theta / \cos^2 \theta} = \frac{1}{\cos \theta} \cdot \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{\cos \theta}{\sin^2 \theta} = \cot \theta \csc \theta$. So the integral becomes

4. Integrate. Recall that $\frac{d}{d\theta} (\csc \theta) = -\cot \theta \csc \theta$. Hence $\int \cot \theta \csc \theta \, d\theta = -\csc \theta + C$. So

5. Back-substitute to $x$. From $x = a \sec \theta$, we have $\sec \theta = \frac{x}{a}$. Draw a right triangle: if $\sec \theta = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{x}{a}$, then adjacent = $a$, hypotenuse = $x$, so opposite = $\sqrt{x^2 - a^2}$. Therefore $\csc \theta = \frac{\text{hypotenuse}}{\text{opposite}} = \frac{x}{\sqrt{x^2 - a^2}}$. Substituting back:

The core idea here is that two different antiderivatives of the same function can differ by a constant. When you evaluate an indefinite integral using two different substitutions, you get expressions that look different but are actually the same up to an additive constant. The question asks for the difference between those two expressions, including their arbitrary constants — and that difference is just a constant.

Let’s work through both methods carefully.

1. First method: $x = \tan \theta$ Substitute $x = \tan \theta$, so $dx = \sec^2 \theta \, d\theta$ and $1 + x^2 = 1 + \tan^2 \theta = \sec^2 \theta$. The integral becomes:

Let $u = \sin \theta$, then $du = \cos \theta \, d\theta$, giving:

Since $\sin \theta = \frac{x}{\sqrt{1+x^2}}$, we get:

2. Second method: $x^2 + 1 = Z$ Substitute $Z = x^2 + 1$, so $dZ = 2x \, dx$ and $x^2 = Z - 1$. The integral becomes:

Since $x \, dx = \frac{dZ}{2}$ and $x^2 = Z-1$, we have:

Integrating:

Substitute back $Z = 1 + x^2$:

3. Compare the two results Write $f(x) + K$ from method 1:

Write $g(x) + C$ from method 2:

Now compute $f(x) - g(x) + K - C$:

Simplify the $x$-terms:

Note $x^4 - 1 = (x^2 - 1)(x^2 + 1)$, so: