Q.Integrate the following function: $\cos 3x$

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 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 concept: recognise the answer, then differentiate to confirm

When an integral like this appears with the answer's shape already given, the fastest and safest route is to differentiate the proposed antiderivative and see whether it reproduces the integrand. The key algebraic hint is the split of the angle:

which is exactly the kind of decomposition the compound-angle formula

is built for.

Step 1 — Differentiate the candidate

Let

By the product rule (and the chain rule on $\sin^{100}x$):

Step 2 — Factor out $100\sin^{99}x$

The bracket is precisely $\sin(100x + x)$:

Step 3 — Integrate

Dividing by 100 and integrating back:

Step 4 — Read off $\lambda$ and $\mu$

The result is written in the form

Comparing term by term:

Step 5 — Reflect

The elegance here is that no substitution or integration by parts is needed at all — the integrand is a derivative in disguise, and the $101 = 100+1$ split is the clue that gives it away.

ANSWER: A

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 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:

Combine:

So $f(x) - g(x) + K - C = \frac14 + (K - C)$.

Since $K$ and $C$ are arbitrary constants, $\frac14 + (K - C)$ is itself an arbitrary constant. Therefore the expression is any constant.