Q.Find $\int \cos 6x \sqrt{1 + \sin 6x}\, dx$

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 key insight is that the expression $\frac{1+\sqrt{3}\cot x}{1-\sqrt{3}\cot x}$ looks like a tangent addition formula in disguise. Recall that $\cot x = \frac{\cos x}{\sin x}$, and $\sqrt{3} = \tan(\pi/3)$. This suggests rewriting the integrand in terms of $\tan$ or $\sin$ and $\cos$ to reveal a simpler structure.

1. Rewrite in terms of sine and cosine Since $\cot x = \frac{\cos x}{\sin x}$, we have:

2. Recognize a tangent addition formula Notice that $\sin x + \sqrt{3}\cos x$ and $\sin x - \sqrt{3}\cos x$ resemble the expansion of $\sin(x \pm \pi/3)$ because:

and similarly,

Multiplying numerator and denominator by 2, we get:

3. Use a trigonometric identity to simplify further The ratio of sines can be expressed using the identity:

Here $A = x + \pi/3$, $B = x - \pi/3$, so $A-B = 2\pi/3$. Thus:

Since $\cos(2\pi/3) = -1/2$ and $\sin(2\pi/3) = \sqrt{3}/2$, we have:

4. Integrate term by term The integral becomes:

Recall that $\int \cot u \, du = \log|\sin u| + c$. So with $u = x - \pi/3$, $du = dx$, we get:

Therefore:

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

We are asked to evaluate

The presence of $\sec x$ and $\tan x$ together suggests using the Weierstrass substitution $t = \tan\frac{x}{2}$. This substitution turns trigonometric integrals into rational ones, which we can handle with partial fractions. The trick is to express $\sec x$ and $\tan x$ in terms of $t$, simplify, and then integrate.

1. Recall the standard Weierstrass substitution formulas:

Also, $dx = \frac{2}{1+t^2} \, dt$.

2. Rewrite the integrand in terms of $t$: The denominator is

Notice $1 + t^2 + 2t = (1+t)^2$, so

Expand:

So denominator becomes $\frac{t^2 + 6t + 5}{1-t^2}$.

The numerator $\sec x$ is $\frac{1+t^2}{1-t^2}$, and $dx = \frac{2}{1+t^2} dt$.

Hence the integral becomes

The factors $(1+t^2)$ and $(1-t^2)$ cancel neatly, leaving

3. Factor the quadratic and use partial fractions: $t^2 + 6t + 5 = (t+1)(t+5)$. So

Multiply through: $2 = A(t+5) + B(t+1)$.

Set $t = -1$: $2 = A(4) \Rightarrow A = \frac{1}{2}$.

Set $t = -5$: $2 = B(-4) \Rightarrow B = -\frac{1}{2}$.

Thus

4. Integrate:

5. Substitute back $t = \tan\frac{x}{2}$:

This matches option (A) exactly.

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.