$\int_0^{\pi/4} \frac{\cos^2 x \sin^2 x}{(\cos^3 x + \sin^3 x)^2} dx$ is equal to Options: (A) 1/6 (B) 1/3 (C) 1/12 (D) 1/9

### Analyzing the Setup The integral provided is: $$I = \int_0^{\pi/4} \frac{\cos^2 x \sin^2 x}{(\cos^3 x + \sin^3 x)^2} dx$$ At first glance, it looks like a tangled mess of powers and trigonometric functions. It is natural to feel intimidated, but in the world of JEE Advanced, intimidation is just a sign that you are about to learn something profound. ### The Power of Observation The first step in any complex problem is to analyze the structure. Look at the denominator: $(\cos^3 x + \sin^3 x)^2$. The power inside is $3$, and the whole expression is squared, giving us an effective degree of $6$. This is our golden clue. In trigonometry, when you see high powers of sine and cosine, your best friend is the $\tan x$ and $\sec x$ relationship. To get there, we need to divide the numerator and the denominator by $\cos^6 x$. ### The Algebraic Alchemy Let us perform the division. For the numerator, we have: $$\frac{\cos^2 x \sin^2 x}{\cos^6 x} = \frac{\sin^2 x}{\cos^4 x} = \tan^2 x \sec^2 x$$ Now, look at the denominator: $$\frac{(\cos^3 x + \sin^3 x)^2}{\cos^6 x} = \left( \frac{\cos^3 x + \sin^3 x}{\cos^3 x} \right)^2 = (1 + \tan^3 x)^2$$ Our integral has now transformed into: $$I = \int_0^{\pi/4} \frac{\tan^2 x \sec^2 x}{(1 + \tan^3 x)^2} dx$$ ### The Magic of Substitution Now, the path forward is clear. We have a function of $\tan x$ and its derivative, $\sec^2 x$, sitting right there in the numerator. Let us set $t = 1 + \tan^3 x$. Differentiating both sides with respect to $x$, we get $dt = 3 \tan^2 x \sec^2 x dx$, which means: $$\tan^2 x \sec^2 x dx = \frac{1}{3} dt$$ We must also update our limits. When $x = 0$, $t = 1 + \tan^3(0) = 1$. When $x = \pi/4$, $t = 1 + \tan^3(\pi/4) = 1 + 1^3 = 2$. Our integral becomes: $$I = \frac{1}{3} \int_1^2 \frac{dt}{t^2}$$ ### Final Calculation We are left with a simple power rule integration. The integral of $t^{-2}$ is $-t^{-1}$. $$I = \frac{1}{3} \left[ -\frac{1}{t} \right]_1^2$$ Evaluating this, we get: $$I = \frac{1}{3} \left( -\frac{1}{2} - (-1) \right) = \frac{1}{3} \left( 1 - \frac{1}{2} \right) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$$ There it is! The complexity has vanished, leaving behind a clean, elegant result of **$1/6$**. Remember, no matter how terrifying an integral looks, there is always a path to simplicity if you look for the underlying symmetry.

JEE Main 2024 (06 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: $\int_{0}^{π /4} \frac{c o s ^{2} x s i n ^{2} x}{( c o s ^{3} x + s i n ^{3} x ) ^{2}} d x$ is equal to

Q: Why do we divide by $\cos^6 x$ specifically? Why not $\sin^6 x$ or some other power?

Great question! We choose $\cos^6 x$ because our goal is to transform the expression into a form involving $\tan x$ and $\sec^2 x$. Since the derivative of $\tan x$ is $\sec^2 x$, this creates a perfect setup for substitution. Dividing by $\cos^6 x$ is the algebraic 'key' that unlocks this specific trigonometric structure.

Q: Wait, why does the denominator become $(1 + \tan^3 x)^2$?

Let's break it down. We have $\frac{(\cos^3 x + \sin^3 x)^2}{\cos^6 x}$. Since $\cos^6 x = (\cos^3 x)^2$, we can rewrite the whole expression as $\left(\frac{\cos^3 x + \sin^3 x}{\cos^3 x}\right)^2$. When you divide each term inside the parenthesis by $\cos^3 x$, you get $\frac{\cos^3 x}{\cos^3 x} + \frac{\sin^3 x}{\cos^3 x}$, which is $1 + \tan^3 x$. Squaring that gives us $(1 + \tan^3 x)^2$.

Select Answer:

Visualized Solution

Analyze the Integral Structure

Given Integral: $I = \int_{0}^{\frac{π}{4}} \frac{c o s ^{2} x s i n ^{2} x}{( c o s ^{3} x + s i n ^{3} x ) ^{2}} d x$
Objective: Simplify the trigonometric expression to make it integrable.
Observation: The total degree of the denominator is $3 \times 2 = 6$ .

Strategy: Divide by $cos^{6} x$

Strategy: Divide numerator and denominator by $cos^{6} x$ .
$I = \int_{0}^{\frac{π}{4}} \frac{\frac{c o s ^{2} x s i n ^{2} x}{c o s ^{6} x}}{\frac{( c o s ^{3} x + s i n ^{3} x ) ^{2}}{c o s ^{6} x}} d x$

Simplify the Numerator

Numerator simplification:
$\frac{c o s ^{2} x s i n ^{2} x}{c o s ^{6} x} = \frac{s i n ^{2} x}{c o s ^{4} x}$
$= \frac{s i n ^{2} x}{c o s ^{2} x} \cdot \frac{1}{c o s ^{2} x} = tan^{2} x sec^{2} x$

Simplify the Denominator

Denominator simplification:
$\frac{( c o s ^{3} x + s i n ^{3} x ) ^{2}}{c o s ^{6} x} = (\frac{c o s ^{3} x + s i n ^{3} x}{c o s ^{3} x})^{2}$
$= (1 + tan^{3} x)^{2}$

The Transformed Integral

New Integral Form:
$I = \int_{0}^{\frac{π}{4}} \frac{t a n ^{2} x s e c ^{2} x}{( 1 + t a n ^{3} x ) ^{2}} d x$

Substitution Method

Let $1 + tan^{3} x = t$
Differentiating both sides with respect to $x$ :
$3 tan^{2} x sec^{2} x d x = d t$
$\Rightarrow tan^{2} x sec^{2} x d x = \frac{1}{3} d t$

Changing the Limits

Change of limits:
When $x = 0, t = 1 + tan^{3} (0) = 1 + 0 = 1$
When $x = \frac{π}{4}, t = 1 + tan^{3} (\frac{π}{4}) = 1 + 1^{3} = 2$

Rewrite and Integrate

Substitute $t$ and limits into the integral:
$I = \frac{1}{3} \int_{1}^{2} \frac{d t}{t ^{2}}$
$I = \frac{1}{3} [- \frac{1}{t}]_{1}^{2}$

Final Calculation

Applying the limits:
$I = \frac{1}{3} (- \frac{1}{2} - (- \frac{1}{1}))$
$I = \frac{1}{3} (1 - \frac{1}{2})$
$I = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$

Conclusion \& Key Takeaway

Final Answer: $\frac{1}{6}$
Key Takeaway: For integrals involving $sin x$ and $cos x$ with high powers, dividing by $cos^{n} x$ to create $tan x$ and $sec^{2} x$ is a powerful strategy.
Challenge: What happens if the denominator was $(cos^{4} x + sin^{4} x)$ ? How would you divide then?

00:00 / 00:00

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The integral provided is:

I = \int_{0}^{π /4} \frac{cos ^{2} x sin ^{2} x}{( cos ^{3} x + sin ^{3} x ) ^{2}} d x

At first glance, it looks like a tangled mess of powers and trigonometric functions. It is natural to feel intimidated, but in the world of JEE Advanced, intimidation is just a sign that you are about to learn something profound.

The Power of Observation

The first step in any complex problem is to analyze the structure. Look at the denominator:

(cos^{3} x + sin^{3} x)^{2}

. The power inside is

3

, and the whole expression is squared, giving us an effective degree of

6

This is our golden clue. In trigonometry, when you see high powers of sine and cosine, your best friend is the

tan x

and

sec x

relationship. To get there, we need to divide the numerator and the denominator by

cos^{6} x

The Algebraic Alchemy

Let us perform the division. For the numerator, we have:

\frac{cos ^{2} x sin ^{2} x}{cos ^{6} x} = \frac{sin ^{2} x}{cos ^{4} x} = tan^{2} x sec^{2} x

Now, look at the denominator:

\frac{( cos ^{3} x + sin ^{3} x ) ^{2}}{cos ^{6} x} = (\frac{cos ^{3} x + sin ^{3} x}{cos ^{3} x})^{2} = (1 + tan^{3} x)^{2}

Our integral has now transformed into:

I = \int_{0}^{π /4} \frac{tan ^{2} x sec ^{2} x}{( 1 + tan ^{3} x ) ^{2}} d x

The Magic of Substitution

Now, the path forward is clear. We have a function of

tan x

and its derivative,

sec^{2} x

, sitting right there in the numerator.

Let us set

t = 1 + tan^{3} x

. Differentiating both sides with respect to

x

, we get

d t = 3 tan^{2} x sec^{2} x d x

, which means:

tan^{2} x sec^{2} x d x = \frac{1}{3} d t

We must also update our limits. When

x = 0

t = 1 + tan^{3} (0) = 1

. When

x = π /4

t = 1 + tan^{3} (π /4) = 1 + 1^{3} = 2

Our integral becomes:

I = \frac{1}{3} \int_{1}^{2} \frac{d t}{t ^{2}}

Final Calculation

We are left with a simple power rule integration. The integral of

t^{- 2}

- t^{- 1}

I = \frac{1}{3} [- \frac{1}{t}]_{1}^{2}

Evaluating this, we get:

I = \frac{1}{3} (- \frac{1}{2} - (- 1)) = \frac{1}{3} (1 - \frac{1}{2}) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}

There it is! The complexity has vanished, leaving behind a clean, elegant result of $1/6$ . Remember, no matter how terrifying an integral looks, there is always a path to simplicity if you look for the underlying symmetry.

Common Traps & FAQs

Why do we divide by

cos^{6} x

specifically? Why not

sin^{6} x

or some other power?

Great question! We choose

cos^{6} x

because our goal is to transform the expression into a form involving

tan x

and

sec^{2} x

. Since the derivative of

tan x

sec^{2} x

, this creates a perfect setup for substitution. Dividing by

cos^{6} x

is the algebraic 'key' that unlocks this specific trigonometric structure.

Wait, why does the denominator become

(1 + tan^{3} x)^{2}

Let's break it down. We have

\frac{( c o s ^{3} x + s i n ^{3} x ) ^{2}}{c o s ^{6} x}

. Since

cos^{6} x = (cos^{3} x)^{2}

, we can rewrite the whole expression as

(\frac{c o s ^{3} x + s i n ^{3} x}{c o s ^{3} x})^{2}

. When you divide each term inside the parenthesis by

cos^{3} x

, you get

\frac{c o s ^{3} x}{c o s ^{3} x} + \frac{s i n ^{3} x}{c o s ^{3} x}

, which is

1 + tan^{3} x

. Squaring that gives us

(1 + tan^{3} x)^{2}

Do I really need to change the limits of integration? Can't I just substitute back at the end?

You *could* substitute back, but changing the limits is much safer and faster for definite integrals. It prevents the common mistake of forgetting to convert the variable back to

x

before applying the bounds. Plus, it's elegant! When

x = 0

t = 1

, and when

x = π /4

t = 2

. It simplifies the final calculation significantly.

Animated Solution for Mathematics - Definite Integration: $\int_{0}^{π /4} \frac{c o s ^{2} x s i n ^{2} x}{( c o s ^{3} x + s i n ^{3} x ) ^{2}} d x$ is equal to

Select Answer:

Visualized Solution

Analyze the Integral Structure

Strategy: Divide by $cos^{6} x$

Simplify the Numerator

Simplify the Denominator

The Transformed Integral

Substitution Method

Changing the Limits

Rewrite and Integrate

Final Calculation

Conclusion \& Key Takeaway

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Power of Observation

The Algebraic Alchemy

The Magic of Substitution

Final Calculation

Similar Questions

The integral $\int \frac{s i n ^{2} x c o s ^{2} x}{( s i n ^{5} x + c o s ^{3} x s i n ^{2} x + s i n ^{3} x c o s ^{2} x + c o s ^{5} x ) ^{2}} d x$ is equal to :

The value of the integral $\int \frac{c o s ^{3} x + c o s ^{5} x}{s i n ^{2} x + s i n ^{4} x} d x$ is

$\int \frac{d x}{c o s x + 3 s i n x}$ equals

$\int \frac{x ^{2} - 1}{x ^{3} 2 x ^{4} - 2 x ^{2} + 1} d x =$

If $\int \frac{c o s x}{s i n ^{3} x ( 1 + s i n ^{6} x ) ^{2/3}} d x = f (x) (1 + sin^{6} x)^{1/3} + c$ , where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ is equal to :

The value of the integral $\int \frac{s i n θ - s i n 2 θ ( s i n ^{3} θ + s i n ^{2} θ + s i n θ )}{2 s i n ^{4} θ + 3 s i n ^{2} θ + 6} \frac{d θ}{1 - c o s 2 θ}$ is (where $c$ is a constant of integration)

Let $I (x) = \int \frac{6}{s i n ^{2} x ( 1 - c o t x ) ^{2}} d x$ . If $I (0) = 3$ , then $I (\frac{π}{12})$ is equal to

Evaluate $\int \frac{( c o s 2 x ) ^{1/2}}{s i n x} d x$

The integral $\int \frac{2 x ^{12} + 5 x ^{9}}{( x ^{5} + x ^{3} + 1 ) ^{3}} d x$ is equal to

If $\int \frac{c o s x d x}{s i n ^{2} x ( 1 + s i n x ) ^{2/3}} = f (x) (1 + sin^{6} x)^{1/ λ} + c$ where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ is equal to :

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫0π/4​(cos3x+sin3x)2cos2xsin2x​dx is equal to

Select Answer:

Visualized Solution

Analyze the Integral Structure

Strategy: Divide by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos6x

Simplify the Numerator

Simplify the Denominator

The Transformed Integral

Substitution Method

Changing the Limits

Rewrite and Integrate

Final Calculation

Conclusion \& Key Takeaway

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Power of Observation

The Algebraic Alchemy

The Magic of Substitution

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Definite Integration: $\int_{0}^{π /4} \frac{c o s ^{2} x s i n ^{2} x}{( c o s ^{3} x + s i n ^{3} x ) ^{2}} d x$ is equal to

Strategy: Divide by $cos^{6} x$

The integral $\int \frac{s i n ^{2} x c o s ^{2} x}{( s i n ^{5} x + c o s ^{3} x s i n ^{2} x + s i n ^{3} x c o s ^{2} x + c o s ^{5} x ) ^{2}} d x$ is equal to :

The value of the integral $\int \frac{c o s ^{3} x + c o s ^{5} x}{s i n ^{2} x + s i n ^{4} x} d x$ is

$\int \frac{d x}{c o s x + 3 s i n x}$ equals

$\int \frac{x ^{2} - 1}{x ^{3} 2 x ^{4} - 2 x ^{2} + 1} d x =$

If $\int \frac{c o s x}{s i n ^{3} x ( 1 + s i n ^{6} x ) ^{2/3}} d x = f (x) (1 + sin^{6} x)^{1/3} + c$ , where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ is equal to :

The value of the integral $\int \frac{s i n θ - s i n 2 θ ( s i n ^{3} θ + s i n ^{2} θ + s i n θ )}{2 s i n ^{4} θ + 3 s i n ^{2} θ + 6} \frac{d θ}{1 - c o s 2 θ}$ is (where $c$ is a constant of integration)

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The integral $\int \frac{2 x ^{12} + 5 x ^{9}}{( x ^{5} + x ^{3} + 1 ) ^{3}} d x$ is equal to

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