The integral ∫ sec² x / (sec x + tan x)^(9/2) dx equals (for some arbitrary constant K) (a) 1/(sec x

Q: Why can't I just use $u = \sec x$ or $u = \tan x$?

That is a very common trap! If you substitute $u = \sec x$, the $\tan x$ term in the denominator remains, and you will be left with a square root of $u^2 - 1$ inside the denominator, which makes the integral significantly harder. The substitution $t = \sec x + \tan x$ is a 'master key' because it collapses the entire complex denominator into a single variable $t$.

Q: How did we get $\sec x = \frac{1}{2}(t + \frac{1}{t})$?

It comes from the identity $\sec^2 x - \tan^2 x = 1$. Factoring this gives $(\sec x + \tan x)(\sec x - \tan x) = 1$. Since we defined $t = \sec x + \tan x$, it follows that $\sec x - \tan x = \frac{1}{t}$. If you add these two equations, the $\tan x$ terms cancel out, leaving $2 \sec x = t + \frac{1}{t}$. It is a beautiful piece of algebraic symmetry!

Q: Why do we factor out $t^{11/2}$ at the very end?

That is a strategic move to match the format of the options provided in the JEE Advanced paper. In competitive exams, the final answer is often presented in a specific algebraic form. Factoring out the term is not just for aesthetics; it is the final step to ensure your result aligns perfectly with the given choices.

Question

The integral $\int \frac{\sec^2 x}{(\sec x + 	an x)^{9/2}} dx$ equals (for some arbitrary constant $K$)

Options:

(A) $\frac{1}{(\sec x + 	an x)^{11/2}} \left\{ \frac{1}{11} - \frac{1}{7} (\sec x + 	an x)^2 ight\} + K$

(B) $\frac{1}{(\sec x + 	an x)^{11/2}} \left\{ \frac{1}{11} - \frac{1}{7} (\sec x + 	an x)^2 ight\} + K$

(C) $\frac{1}{(\sec x + 	an x)^{11/2}} \left\{ \frac{1}{11} + \frac{1}{7} (\sec x + 	an x)^2 ight\} + K$

(D) $\frac{1}{(\sec x + 	an x)^{11/2}} \left\{ \frac{1}{11} + \frac{1}{7} (\sec x + 	an x)^2 ight\} + K$

Sigma Percentile · Accepted Answer

### The Art of the Substitution: Taming the Trigonometric Beast

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of calculus. We are looking at the integral:

$$I = \int \frac{\sec^2 x}{(\sec x + 	an x)^{9/2}} dx$$

When you see an expression like this in a JEE Advanced paper, your heart might skip a beat. You see the $\sec^2 x$ in the numerator, and you see that massive, intimidating power of $9/2$ in the denominator.

Most students freeze here. They try to convert everything to $\sin x$ and $\cos x$, which leads to a swamp of algebra. But you? You are going to look for the hidden structure.

### Phase 1: The Intuition

In calculus, especially in JEE, integration is often about recognizing patterns. Look at the denominator: $(\sec x + 	an x)$. Now, ask yourself: what is the derivative of this expression?

Recall that $\frac{d}{dx}(\sec x) = \sec x 	an x$ and $\frac{d}{dx}(	an x) = \sec^2 x$. If we differentiate $t = \sec x + 	an x$, we get:

$$\frac{dt}{dx} = \sec x 	an x + \sec^2 x$$

Look at that! If we factor out $\sec x$, we get $\sec x (	an x + \sec x)$. This is exactly our substitution $t$ multiplied by $\sec x$.

This is the 'Aha!' moment. The numerator contains the derivative of the denominator, albeit with a slight complication. This is the signal to proceed with substitution.

### Phase 2: The Algebraic Bridge

We have defined $t = \sec x + 	an x$. Our differential is $\frac{dt}{dx} = \sec x \cdot t$, which implies $\sec x dx = \frac{dt}{t}$.

But wait—our original integral has $\sec^2 x dx$. We have accounted for $\sec x dx$, but we have one $\sec x$ left over. How do we express this remaining $\sec x$ in terms of $t$?

This is where the fundamental identity $\sec^2 x - 	an^2 x = 1$ becomes our best friend. We factor it as a difference of squares:

$$(\sec x + 	an x)(\sec x - 	an x) = 1$$

Since $\sec x + 	an x = t$, we immediately see that $\sec x - 	an x = \frac{1}{t}$. Now we have a system of two equations:

1) $\sec x + 	an x = t$
2) $\sec x - 	an x = \frac{1}{t}$

Adding these two equations eliminates the $	an x$ term, giving us $2 \sec x = t + \frac{1}{t}$, or simply $\sec x = \frac{1}{2}(t + \frac{1}{t})$. We have successfully bridged the gap between the trigonometric world and the algebraic world.

### Phase 3: The Transformation

Now, let us rewrite the integral. We split $\sec^2 x dx$ into $\sec x \cdot (\sec x dx)$. Substituting our expressions, the integral becomes:

$$I = \int \frac{\frac{1}{2}(t + \frac{1}{t}) \cdot \frac{dt}{t}}{t^{9/2}}$$

Look at the beauty of this transformation. The trigonometric functions have vanished, replaced by a clean, manageable algebraic expression. Let us simplify the integrand:

$$I = \frac{1}{2} \int \frac{t + t^{-1}}{t \cdot t^{9/2}} dt = \frac{1}{2} \int \frac{t + t^{-1}}{t^{11/2}} dt$$

Distributing the division, we get:

$$I = \frac{1}{2} \int (t^{-9/2} + t^{-13/2}) dt$$

### Phase 4: The Final Integration

We are in the home stretch. Applying the power rule $\int t^n dt = \frac{t^{n+1}}{n+1}$, we get:

$$I = \frac{1}{2} \left[ \frac{t^{-7/2}}{-7/2} + \frac{t^{-11/2}}{-11/2} ight] + K$$

Simplifying the coefficients, the $2$ in the denominator cancels with the $2$ in the numerator of the fractions:

$$I = -\frac{1}{7} t^{-7/2} - \frac{1}{11} t^{-11/2} + K$$

To match the options, we factor out $-\frac{1}{t^{11/2}}$:

$$I = -\frac{1}{t^{11/2}} \left[ \frac{1}{11} + \frac{t^2}{7} ight] + K$$

Finally, substituting $t = \sec x + 	an x$ back into the equation, we arrive at the solution. This problem wasn't about brute force; it was about recognizing the symmetry between $\sec x$ and $	an x$. Keep this logic in your toolkit—it will serve you well in the exam hall.

Animated Solution for Mathematics - Indefinite Integration: The integral $\int \frac{s e c ^{2} x}{( s e c x + t a n x ) ^{9/2}} d x$ equals (for some arbitrary constant $K$ )

Select Answer:

Visualized Solution

Analyze the Integral Structure

Define the Substitution $t$

Differentiate the Substitution

Isolate the Differential

Relate $sec x$ and $t$

Express $sec x$ in terms of $t$

Prepare the Integral for Substitution

Rewrite the Original Integral

Substitute into the Integral

Simplify the Integrand

Separate into Power Terms

Integrate using Power Rule

Simplify the Coefficients

Factor out the Common Term

Final Substitution and Summary

The Sigma Insight: Integration by Substitution

The Art of the Substitution

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Indefinite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫(secx+tanx)9/2sec2x​dx equals (for some arbitrary constant K)