If ∫ 5 tan x / (tan x - 2) dx = x + a ln |sin x - 2 cos x| + k, then a is equal to (a) -1 (b) -2 (c)

Q: Why can't I just substitute $u = \tan x$?

That is a fantastic question! If you set $u = \tan x$, you would need $du = \sec^2 x dx$ to be present in the integral to complete the substitution. Since we do not have a $\sec^2 x$ term, that path leads to a dead end. We need a method that handles the linear combination of sine and cosine directly, which is why we convert to $\sin x$ and $\cos x$ first.

Q: How do I know when to use the $A(\text{Denom}) + B(\text{Denom}')$ method?

Look for the structure $\int \frac{p \sin x + q \cos x}{r \sin x + s \cos x} dx$. Whenever you see a linear combination of sine and cosine in both the numerator and denominator, this method is your 'golden key'. It is designed specifically to separate the integral into a part that integrates to $x$ and a part that integrates to $\ln|\text{denominator}|$. It is a standard tool in your JEE toolkit for these types of problems.

Q: Wait, why do we reject the negative root or worry about the absolute value?

In the term $\ln |\sin x - 2 \cos x|$, the absolute value is crucial because the logarithm function is only defined for positive arguments. Since $\sin x - 2 \cos x$ can take negative values depending on the value of $x$, the absolute value ensures our mathematical model remains valid across the entire domain. Never forget the absolute value in logarithmic integrals!

Question

If $\int \frac{5 	an x}{	an x - 2} dx = x + a \ln |\sin x - 2 \cos x| + k$, then $a$ is equal to

Options:

(A) $-1$

(B) $-2$

(C) $1$

(D) $2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are tackling an integral that might look like a tangled mess of trigonometric functions at first glance:

$$I = \int \frac{5 	an x}{	an x - 2} dx$$

When you see an expression like this, it is natural to feel a moment of hesitation. But remember, in the world of JEE Advanced, complexity is often just a mask for a beautiful, underlying structure. Our goal is to peel back that mask.

### The Great Simplification

The first step in our journey is to simplify the landscape. We have $	an x$ everywhere, which is a ratio of $\sin x$ and $\cos x$. By substituting $	an x = \frac{\sin x}{\cos x}$, we transform our integral into:

$$I = \int \frac{5 \left( \frac{\sin x}{\cos x} ight)}{\frac{\sin x}{\cos x} - 2} dx$$

To clear the clutter, we multiply the numerator and the denominator by $\cos x$. Suddenly, the chaos vanishes, and we are left with a much cleaner, more elegant form:

$$I = \int \frac{5 \sin x}{\sin x - 2 \cos x} dx$$

This is the moment where the problem shifts from 'impossible' to 'solvable'.

### The Linear Combination Strategy

Now, we face a fraction where both the numerator and the denominator are linear combinations of $\sin x$ and $\cos x$. This is a classic scenario in calculus.

We want to express the numerator, $5 \sin x$, as a combination of the denominator, $(\sin x - 2 \cos x)$, and its derivative. Let us calculate that derivative first:

$$\frac{d}{dx}(\sin x - 2 \cos x) = \cos x + 2 \sin x$$

So, we set up our identity:

$$5 \sin x = A(\sin x - 2 \cos x) + B(\cos x + 2 \sin x)$$

This is the 'secret key' to the problem. By finding the constants $A$ and $B$, we can split this single, difficult integral into two manageable pieces.

### Solving for the Constants

We expand the right side of our identity:

$$5 \sin x = (A + 2B) \sin x + (-2A + B) \cos x$$

By comparing the coefficients of $\sin x$ and $\cos x$ on both sides, we get a system of linear equations:

$$A + 2B = 5$$
$$-2A + B = 0$$

From the second equation, we see that $B = 2A$. Substituting this into the first equation, we get $A + 2(2A) = 5$, which simplifies to $5A = 5$, giving us **$A = 1$**. Consequently, **$B = 2$**.

### The Final Integration

With $A = 1$ and $B = 2$, our integral becomes:

$$I = \int \frac{1(\sin x - 2 \cos x) + 2(\cos x + 2 \sin x)}{\sin x - 2 \cos x} dx$$

We split this into two parts:

$$I = \int 1 dx + 2 \int \frac{\cos x + 2 \sin x}{\sin x - 2 \cos x} dx$$

The first part is trivial: $\int 1 dx = x$. The second part is a standard integral of the form $\int \frac{f'(x)}{f(x)} dx$, which evaluates to $\ln |f(x)|$.

Thus, the second part becomes $2 \ln |\sin x - 2 \cos x|$. Adding the constant of integration $k$, we arrive at our final result:

$$I = x + 2 \ln |\sin x - 2 \cos x| + k$$

Comparing this to the given form $x + a \ln |\sin x - 2 \cos x| + k$, we immediately see that **$a = 2$**. You have successfully navigated the complexity and emerged victorious.

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{5 t a n x}{t a n x - 2} d x = x + a ln ∣ sin x - 2 cos x ∣ + k$ , then $a$ is equal to

Select Answer:

Visualized Solution

Orienting the Integral $I = \int \frac{5 t a n x}{t a n x - 2} d x$

Expressing in $sin x$ and $cos x$

Algebraic Simplification

The Linear Combination Method

Setting up the Identity

Comparing Coefficients

Solving for $A$ and $B$

Splitting the Integral

Final Integration and Comparison

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Great Simplification

The Linear Combination Strategy

Solving for the Constants

The Final Integration

Similar Questions

If $\int \frac{t a n x}{1 + t a n x + t a n ^{2} x} d x = x - \frac{K}{A} tan^{- 1} (\frac{K t a n x + 1}{A}) + C$ , (C is a constant of integration), then the ordered pair (K, A) is equal to :

If $\int \frac{c o s θ}{5 + 7 s i n θ - 2 c o s ^{2} θ} d θ = A lo g_{e} ∣ B (θ) ∣ + C$ where $C$ is a constant of integration, then $\frac{B ( θ )}{A}$ can be:

If $\int \frac{d x}{( x ^{2} - 2 x + 10 ) ^{2}} = A (tan^{- 1} (\frac{x - 1}{3}) + \frac{f ( x )}{x ^{2} - 2 x + 10}) + C$ , where $C$ is a constant of integration, then :

If $\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) d x = g (x) + C,$ where C is the constant of integration, then $g (\frac{1}{2})$ equals:

The integral $\int \frac{( 1 - \frac{1}{3} ) ( c o s x - s i n x )}{1 + \frac{2}{3} s i n 2 x} d x$ is equal to

Evaluate $\int \frac{s i n x}{s i n x - c o s x} d x$

$\int {\frac{l o g x - 1}{1 + ( l o g x ) ^{2}}}^{2} d x$ is equal to

$\int \frac{d x}{c o s x - s i n x}$ is equal to

If $\int \frac{2 x ^{2} + 5 x + 9}{x ^{2} + x + 1} d x = x x^{2} + x + 1 + α x^{2} + x + 1 + β lo g_{e} ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣ + C .$ where C is the constant of integration, then $α + 2 β$ is equal to

Animated Solution for Mathematics - Indefinite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫tanx−25tanx​dx=x+aln∣sinx−2cosx∣+k, then a is equal to