If the integral $525 \int_{0}^{\pi/2} \sin 2x \cos^{11/2} x (1 + \cos^{5/2} x)^{1/2} dx$ is equal to $(n\sqrt{2} - 64)$, then $n$ is equal to _______.

### Analyzing the Setup Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an integral; we are peeling back the layers of a mathematical onion. At first glance, the expression $I = \int_{0}^{\pi/2} \sin 2x \cos^{11/2} x (1 + \cos^{5/2} x)^{1/2} dx$ might look like a chaotic mess of fractional powers and trigonometric functions. In the world of advanced calculus, complexity is often just a mask for hidden symmetry. ### The Simplification Our first move is to bring order to the chaos. We see a $\sin 2x$ term, and our intuition screams for the double-angle identity. By applying $\sin 2x = 2 \sin x \cos x$, we transform the integrand into: $$I = \int_{0}^{\pi/2} (2 \sin x \cos x) \cos^{11/2} x (1 + \cos^{5/2} x)^{1/2} dx$$ By combining the powers of cosine, $\cos x \cdot \cos^{11/2} x$, we arrive at $\cos^{13/2} x$. Now, the integral simplifies to: $$I = 2 \int_{0}^{\pi/2} \sin x \cos^{13/2} x (1 + \cos^{5/2} x)^{1/2} dx$$ ### The First Transformation We face fractional powers, which act as obstacles in our path. To clear them, we perform a strategic substitution. Let $\cos x = t^2$. This implies that $-\sin x dx = 2t dt$. As we change our variable from $x$ to $t$, we must also update our limits: when $x=0$, $t=1$; when $x=\pi/2$, $t=0$. The negative sign from the differential allows us to flip the limits back to the natural order of $[0, 1]$. The integral becomes: $$I = 4 \int_{0}^{1} t^{13} \sqrt{1 + t^5} \cdot t dt = 4 \int_{0}^{1} t^{14} \sqrt{1 + t^5} dt$$ ### The Final Key We are almost there. We have $\sqrt{1 + t^5}$ staring us in the face. Let's define $1 + t^5 = k^2$. This is the master key. Differentiating gives $5t^4 dt = 2k dk$, or $t^4 dt = \frac{2}{5} k dk$. We also know $t^5 = k^2 - 1$, so $t^{10} = (k^2 - 1)^2$. By splitting $t^{14}$ into $t^{10} \cdot t^4$, we rewrite the integral in terms of $k$: $$I = 4 \int_{1}^{\sqrt{2}} (k^2 - 1)^2 \cdot k \cdot \left(\frac{2k}{5} dk\right) = \frac{8}{5} \int_{1}^{\sqrt{2}} (k^6 - 2k^4 + k^2) dk$$ ### The Victory Lap We have arrived at the finish line. The integration is now trivial, a simple application of the power rule: $$I = \frac{8}{5} \left[ \frac{k^7}{7} - \frac{2k^5}{5} + \frac{k^3}{3} \right]_{1}^{\sqrt{2}}$$ Evaluating this at the limits $\sqrt{2}$ and $1$ requires careful arithmetic. After calculating the values, we find: $$I = \frac{176\sqrt{2} - 64}{525}$$ Comparing this to our target expression $525 I = n\sqrt{2} - 64$, we see clearly that **$n = 176$**. Every complex problem is just a series of simple truths waiting to be revealed.

JEE Main 2024 (31 Jan Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: If the integral $525 \int_{0}^{π /2} sin 2 x cos^{11/2} x (1 + cos^{5/2} x)^{1/2} d x$ is equal to $(n 2 - 64)$ , then $n$ is equal to _______.

Q: Why did we choose $\cos x = t^2$ instead of just $\cos x = t$?

Great observation! If you set $\cos x = t$, you would be left with $\sqrt{1 + t^{5/2}}$, which is still difficult to integrate. By choosing $\cos x = t^2$, we effectively clear the fractional power of $1/2$ inside the square root, turning the expression into $\sqrt{1 + t^5}$, which is much easier to handle with a second substitution.

Q: I got lost when changing the limits. Why did the limits flip from $[1, 0]$ to $[0, 1]$?

Don't worry, that's a classic trap! When we substituted $\cos x = t^2$, the differential $-\sin x dx = 2t dt$ introduced a negative sign. We used that negative sign to swap the limits of integration back to the standard $[0, 1]$ order. It's a standard property of definite integrals: $\int_a^b f(x) dx = -\int_b^a f(x) dx$.

Enter Numerical Value:

Visualized Solution

Visualizing the Integral

Given Integral: $I = \int_{0}^{\frac{π}{2}} sin 2 x cos^{\frac{11}{2}} x (1 + cos^{\frac{5}{2}} x)^{\frac{1}{2}} d x$
Target: Find $n$ such that $525 I = n 2 - 64$
The graph shows the integrand function over the interval $[0, \frac{π}{2}]$ .

Expanding $sin 2 x$

Use the double angle identity: $sin 2 x = 2 sin x cos x$
Substitute into the integral:
$I = \int_{0}^{\frac{π}{2}} (2 sin x cos x) cos^{\frac{11}{2}} x (1 + cos^{\frac{5}{2}} x)^{\frac{1}{2}} d x$

Simplifying Cosine Powers

Combine the powers of cosine: $cos x \cdot cos^{\frac{11}{2}} x = cos^{\frac{13}{2}} x$
The simplified integral becomes:
$I = 2 \int_{0}^{\frac{π}{2}} sin x cos^{\frac{13}{2}} x (1 + cos^{\frac{5}{2}} x)^{\frac{1}{2}} d x$

First Substitution: $cos x = t^{2}$

To handle the fractional powers, let $cos x = t^{2}$
Differentiating both sides: $- sin x d x = 2 t d t$
Therefore, $sin x d x = - 2 t d t$

Changing the Limits

Lower limit: When $x = 0$ , $t^{2} = cos 0 = 1 \Rightarrow t = 1$
Upper limit: When $x = \frac{π}{2}$ , $t^{2} = cos (\frac{π}{2}) = 0 \Rightarrow t = 0$
The integral limits change from $[0, \frac{π}{2}]$ to $[1, 0]$ .

Transforming the Integral

Substitute $t$ into the integral:
$I = 2 \int_{1}^{0} (t^{2})^{\frac{13}{2}} (1 + (t^{2})^{\frac{5}{2}})^{\frac{1}{2}} (- 2 t d t)$
Reversing limits absorbs the negative sign:
$I = 4 \int_{0}^{1} t^{13} 1 + t^{5} \cdot t d t = 4 \int_{0}^{1} t^{14} 1 + t^{5} d t$

Second Substitution: $1 + t^{5} = k^{2}$

Let $1 + t^{5} = k^{2}$ to eliminate the square root.
Differentiating: $5 t^{4} d t = 2 k d k \Rightarrow t^{4} d t = \frac{2 k}{5} d k$
Also, $t^{5} = k^{2} - 1$ , so $t^{10} = (k^{2} - 1)^{2}$

New Limits for $k$

Lower limit: When $t = 0$ , $k^{2} = 1 + 0 = 1 \Rightarrow k = 1$
Upper limit: When $t = 1$ , $k^{2} = 1 + 1 = 2 \Rightarrow k = 2$
The new limits for $k$ are $[1, 2]$ .

Integral in terms of $k$

Rewrite $t^{14} d t$ as $t^{10} \cdot t^{4} d t$ :
$I = 4 \int_{1}^{2} (k^{2} - 1)^{2} \cdot k \cdot (\frac{2 k}{5} d k)$
Expand the square and multiply:
$I = \frac{8}{5} \int_{1}^{2} (k^{4} - 2 k^{2} + 1) k^{2} d k = \frac{8}{5} \int_{1}^{2} (k^{6} - 2 k^{4} + k^{2}) d k$

Atomic Integration

Integrate each term using the power rule:
$I = \frac{8}{5} [\frac{k ^{7}}{7} - \frac{2 k ^{5}}{5} + \frac{k ^{3}}{3}]_{1}^{2}$

Evaluating Upper Limit

Substitute $k = 2$ :
$\frac{8}{5} (\frac{8 2}{7} - \frac{8 2}{5} + \frac{2 2}{3})$
Take LCM as $105$ :
$\frac{8}{5} \cdot \frac{120 2 - 168 2 + 70 2}{105} = \frac{176 2}{525}$

Evaluating Lower Limit

Substitute $k = 1$ :
$\frac{8}{5} (\frac{1}{7} - \frac{2}{5} + \frac{1}{3})$
Take LCM as $105$ :
$\frac{8}{5} \cdot \frac{15 - 42 + 35}{105} = \frac{64}{525}$
Total Integral $I = \frac{176 2 - 64}{525}$

Final Comparison

We have $I = \frac{176 2 - 64}{525}$
Multiply by $525$ : $525 I = 1762 - 64$
Compare with the given expression $n 2 - 64$
Therefore, $n = 176$ .

00:00 / 00:00

The Sigma Insight: Evaluation of Special Integral Forms

Solution Diagram

Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an integral; we are peeling back the layers of a mathematical onion.

At first glance, the expression

I = \int_{0}^{π /2} sin 2 x cos^{11/2} x (1 + cos^{5/2} x)^{1/2} d x

might look like a chaotic mess of fractional powers and trigonometric functions. In the world of advanced calculus, complexity is often just a mask for hidden symmetry.

The Simplification

Our first move is to bring order to the chaos. We see a

sin 2 x

term, and our intuition screams for the double-angle identity. By applying

sin 2 x = 2 sin x cos x

, we transform the integrand into:

I = \int_{0}^{π /2} (2 sin x cos x) cos^{11/2} x (1 + cos^{5/2} x)^{1/2} d x

By combining the powers of cosine,

cos x \cdot cos^{11/2} x

, we arrive at

cos^{13/2} x

. Now, the integral simplifies to:

I = 2 \int_{0}^{π /2} sin x cos^{13/2} x (1 + cos^{5/2} x)^{1/2} d x

The First Transformation

We face fractional powers, which act as obstacles in our path. To clear them, we perform a strategic substitution. Let

cos x = t^{2}

This implies that

- sin x d x = 2 t d t

. As we change our variable from

x

t

, we must also update our limits: when

x = 0

t = 1

; when

x = π /2

t = 0

The negative sign from the differential allows us to flip the limits back to the natural order of

[0, 1]

. The integral becomes:

I = 4 \int_{0}^{1} t^{13} 1 + t^{5} \cdot t d t = 4 \int_{0}^{1} t^{14} 1 + t^{5} d t

The Final Key

We are almost there. We have

1 + t^{5}

staring us in the face. Let's define

1 + t^{5} = k^{2}

. This is the master key.

Differentiating gives

5 t^{4} d t = 2 k d k

, or

t^{4} d t = \frac{2}{5} k d k

. We also know

t^{5} = k^{2} - 1

, so

t^{10} = (k^{2} - 1)^{2}

. By splitting

t^{14}

into

t^{10} \cdot t^{4}

, we rewrite the integral in terms of

k

I = 4 \int_{1}^{2} (k^{2} - 1)^{2} \cdot k \cdot (\frac{2 k}{5} d k) = \frac{8}{5} \int_{1}^{2} (k^{6} - 2 k^{4} + k^{2}) d k

The Victory Lap

We have arrived at the finish line. The integration is now trivial, a simple application of the power rule:

I = \frac{8}{5} [\frac{k ^{7}}{7} - \frac{2 k ^{5}}{5} + \frac{k ^{3}}{3}]_{1}^{2}

Evaluating this at the limits

2

and

1

requires careful arithmetic. After calculating the values, we find:

I = \frac{176 2 - 64}{525}

Comparing this to our target expression

525 I = n 2 - 64

, we see clearly that $n = 176$ . Every complex problem is just a series of simple truths waiting to be revealed.

Common Traps & FAQs

Why did we choose

cos x = t^{2}

instead of just

cos x = t

Great observation! If you set

cos x = t

, you would be left with

1 + t^{5/2}

, which is still difficult to integrate. By choosing

cos x = t^{2}

, we effectively clear the fractional power of

1/2

inside the square root, turning the expression into

1 + t^{5}

, which is much easier to handle with a second substitution.

Is there a way to solve this using Beta functions?

While Beta functions are powerful, they are usually designed for integrals of the form

\int_{0}^{1} x^{m} (1 - x)^{n} d x

. Because our expression involves

1 + t^{5}

, it does not fit the standard Beta function template directly. Algebraic substitution is the most robust and elegant path here.

I got lost when changing the limits. Why did the limits flip from

[1, 0]

[0, 1]

Don't worry, that's a classic trap! When we substituted

cos x = t^{2}

, the differential

- sin x d x = 2 t d t

introduced a negative sign. We used that negative sign to swap the limits of integration back to the standard

[0, 1]

order. It's a standard property of definite integrals:

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x

Why did we split

t^{14} d t

into

t^{10} \cdot t^{4} d t

That is the 'secret sauce' of this problem! We knew from our second substitution

1 + t^{5} = k^{2}

that

5 t^{4} d t = 2 k d k

. By isolating

t^{4} d t

, we created the exact differential needed to complete the substitution, while

t^{10}

became

(k^{2} - 1)^{2}

, which is a simple polynomial.

How do I know when to stop simplifying and start integrating?

You stop when your integrand is a polynomial or a simple power function. Once we reached

\frac{8}{5} \int (k^{6} - 2 k^{4} + k^{2}) d k

, the 'physics' of the problem was done, and it became a simple exercise in calculus. Always aim to reduce the complexity until the integral looks like a standard power rule problem.

Animated Solution for Mathematics - Definite Integration: If the integral $525 \int_{0}^{π /2} sin 2 x cos^{11/2} x (1 + cos^{5/2} x)^{1/2} d x$ is equal to $(n 2 - 64)$ , then $n$ is equal to _______.

Enter Numerical Value:

Visualized Solution

Visualizing the Integral

Expanding $sin 2 x$

Simplifying Cosine Powers

First Substitution: $cos x = t^{2}$

Changing the Limits

Transforming the Integral

Second Substitution: $1 + t^{5} = k^{2}$

New Limits for $k$

Integral in terms of $k$

Atomic Integration

Evaluating Upper Limit

Evaluating Lower Limit

Final Comparison

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Simplification

The First Transformation

The Final Key

The Victory Lap

Similar Questions

If $\int_{0}^{1} (x^{21} + x^{14} + x^{7}) (2 x^{21} + 3 x^{14} + 6 x^{7})^{1/7} d x = \frac{1}{l} (11)^{m / n}$ where $l, m, n \in N$ , $m$ and $n$ are coprime then $l + m + n$ 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

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If $\int \frac{1}{a ^{2} s i n ^{2} x + b ^{2} c o s ^{2} x} d x = \frac{1}{12} tan^{- 1} (3 tan x) + constant$ , then the maximum value of $a sin x + b cos x$ , is :

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the integral 525∫0π/2​sin2xcos11/2x(1+cos5/2x)1/2dx is equal to (n2​−64), then n is equal to _______.

Enter Numerical Value:

Visualized Solution

Visualizing the Integral

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2x

Simplifying Cosine Powers

First Substitution: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosx=t2

Changing the Limits

Transforming the Integral

Second Substitution: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1+t5=k2

New Limits for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Integral in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Atomic Integration

Evaluating Upper Limit

Evaluating Lower Limit

Final Comparison

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Simplification

The First Transformation

The Final Key

The Victory Lap

Similar Questions

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Animated Solution for Mathematics - Definite Integration: If the integral $525 \int_{0}^{π /2} sin 2 x cos^{11/2} x (1 + cos^{5/2} x)^{1/2} d x$ is equal to $(n 2 - 64)$ , then $n$ is equal to _______.

Expanding $sin 2 x$

First Substitution: $cos x = t^{2}$

Second Substitution: $1 + t^{5} = k^{2}$

New Limits for $k$

Integral in terms of $k$

If $\int_{0}^{1} (x^{21} + x^{14} + x^{7}) (2 x^{21} + 3 x^{14} + 6 x^{7})^{1/7} d x = \frac{1}{l} (11)^{m / n}$ where $l, m, n \in N$ , $m$ and $n$ are coprime then $l + m + n$ 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

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

If $I (x) = \int e^{s i n^{2} x} (cos x sin 2 x - sin x) d x$ and $I (0) = 1$ , then $I (\frac{π}{3})$ is equal to

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

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:

If $\int \frac{1}{a ^{2} s i n ^{2} x + b ^{2} c o s ^{2} x} d x = \frac{1}{12} tan^{- 1} (3 tan x) + constant$ , then the maximum value of $a sin x + b cos x$ , is :

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: