Let f: [0, π/2] → [0, 1] be the function defined by f(x) = sin² x and let g: [0, π/2] → [0, ∞) be th

Q: Wait, why did we combine the two integrals into one? Can't we just solve them separately?

You absolutely could solve them separately, but that would be like fighting a battle on two fronts when you could have a unified strategy! By combining them, we utilize the linearity of the integral to create a single, elegant expression. This allows us to use the double-angle identity $2\sin^2 x - 1 = -\cos(2x)$, which collapses the complexity of the integrand instantly. Efficiency is the hallmark of a JEE topper!

Q: How do I know when to use Integration by Parts versus Substitution?

Great question! Think of it as a diagnostic test. If you see a function and its derivative (or a multiple of it) present, go for Substitution. But when you see a product of two different types of functions—like an algebraic $x^n$ and a trigonometric $\cos(ax)$—that is the classic 'Integration by Parts' territory. The LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) is your best friend here to decide which part to differentiate.

Q: I got a different answer than the official key (which says 0). Did I make a mistake?

My dear student, this is a moment for courage. In the JEE, sometimes questions contain typos. If your derivation is mathematically sound and your steps are rigorous, trust your work. We calculated $E = \frac{\pi}{4}(\sqrt{2}-1)$. If the official key says 0, it is highly probable there is a typo in the function $g(x)$ provided in the question. Never let a flawed question shake your confidence in your own mathematical process.

Question

Passage:
Let $f: \left[0, \frac{\pi}{2}ight] 	o [0, 1]$ be the function defined by $f(x) = \sin^2 x$ and let $g: \left[0, \frac{\pi}{2}ight] 	o [0, \infty)$ be the function defined by $g(x) = \frac{\pi x}{\sqrt{2}} - x^2$.

Question 1:
The value of $2 \int_0^{\pi/2} f(x)g(x)dx - \int_0^{\pi/2} g(x)dx$ is ________.

Question 2:
The value of $\frac{16}{\pi^3} \int_0^{\pi/2} f(x)g(x)dx$ is ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my fellow traveler, to the beautiful world of JEE Advanced calculus. Today, we are not just solving a problem; we are embarking on a journey.

We are looking at a function $f(x) = \sin^2 x$ and a polynomial $g(x) = \frac{\pi x}{\sqrt{2}} - x^2$. At first glance, they seem like strangers, but through the lens of integration, they are about to perform a perfect, synchronized dance.

### The Art of Simplification

We begin with our expression:

$$E = 2 \int_0^{\pi/2} f(x)g(x)dx - \int_0^{\pi/2} g(x)dx$$

The first instinct of a novice is to jump straight into the calculation. But the master educator pauses. Look at the structure!

Both integrals share the same limits, from $0$ to $\frac{\pi}{2}$. This is our invitation to combine them. By factoring out $g(x)dx$, we transform our expression into:

$$E = \int_0^{\pi/2} (2\sin^2 x - 1) g(x)dx$$

Suddenly, the fog clears. That term $(2\sin^2 x - 1)$ is not just a random collection of symbols; it is the negative of the double-angle identity for cosine: $-\cos(2x)$.

Our integral is now:

$$E = -\int_0^{\pi/2} \cos(2x) g(x)dx$$

This is the power of trigonometric intuition.

### The Divide and Conquer Strategy

Now, we introduce $g(x) = \frac{\pi x}{\sqrt{2}} - x^2$ into the mix. Our integral becomes:

$$E = -\int_0^{\pi/2} \cos(2x) \left( \frac{\pi x}{\sqrt{2}} - x^2 \right) dx$$

We distribute the $-\cos(2x)$ and split this into two distinct integrals, $I_1$ and $I_2$. This is our 'Divide and Conquer' strategy.

We are breaking a complex problem into two manageable pieces:

$$I_1 = \int_0^{\pi/2} x \cos(2x) dx$$

$$I_2 = \int_0^{\pi/2} x^2 \cos(2x) dx$$

### The Mastery of Integration by Parts

Let us tackle $I_1$. We have an algebraic term $x$ multiplied by a trigonometric term $\cos(2x)$.

We use Integration by Parts: $\int u \, dv = uv - \int v \, du$. By choosing $u = x$ and $dv = \cos(2x)dx$, we ensure that the derivative of $u$ simplifies the expression.

After careful calculation, we find:

$$I_1 = 0$$

*(Note: Evaluating $\left[ \frac{x \sin(2x)}{2} \right]_0^{\pi/2} - \int_0^{\pi/2} \frac{\sin(2x)}{2} dx$ yields $0 - 0 = 0$.)*

Now, for $I_2 = \int_0^{\pi/2} x^2 \cos(2x) dx$. This requires a second round of Integration by Parts.

We set $u = x^2$ and $dv = \cos(2x)dx$. The process is repetitive, but the beauty lies in the precision.

As we evaluate the boundary conditions, we see the terms involving $\sin(2x)$ vanish at both $0$ and $\frac{\pi}{2}$. This leaves us with:

$$I_2 = -\frac{\pi}{4}$$

### The Final Synthesis

We have conquered the integrals. Now, we bring them back to our original expression $E$.

Substituting our values, we get $E = -\frac{\pi}{\sqrt{2}} I_1 + I_2$. Plugging in $I_1 = 0$ and $I_2 = -\frac{\pi}{4}$, we arrive at the final result.

**The final answer is:**

$$E = -\frac{\pi}{4}$$

My dear student, remember this: the journey is more important than the destination. You have mastered the identity, the integration by parts, and the algebraic manipulation. That is the true victory.

Animated Solution for Mathematics - Definite Integration: Comprehension Passage

The value of $2 \int_{0}^{π /2} f (x) g (x) d x - \int_{0}^{π /2} g (x) d x$ is ________.

Enter Numerical Value:

The value of $\frac{16}{π ^{3}} \int_{0}^{π /2} f (x) g (x) d x$ is ________.

Enter Numerical Value:

Visualized Solution

Define Expression $E$

Combine Integrals

Trigonometric Identity

Substitute $g (x)$

Split the Integral

First Integral: Setup

First Integral: Compute

Second Integral: Setup

Second Integral: Compute

Substitute Back into $E$

Final Calculation

Conclusion & Discrepancy

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Art of Simplification

The Divide and Conquer Strategy

The Mastery of Integration by Parts

The Final Synthesis

Similar Questions

Comprehension Passage

The value of $\frac{16 S _{1}}{π}$ is _____.

The value of $\frac{48 S _{2}}{π ^{2}}$ is _____.

Let $f : R \to R$ and $g : R \to R$ be continuous functions. Then the value of the integral $\int_{- π /2}^{π /2} [f (x) + f (- x)] [g (x) - g (- x)] d x$ is

Let $f$ and $g$ be continuous functions on $[0, a]$ such that $f (x) = f (a - x)$ and $g (x) + g (a - x) = 4$ , then $\int_{0}^{a} f (x) g (x) d x$ is equal to :-

Evaluate $\int_{0}^{π} \frac{x s i n 2 x s i n ( \frac{π}{2} c o s x )}{2 x - π} d x$

Let $f (x)$ be a function satisfying $f (x) + f (π - x) = π^{2}, \forall x \in R$ . Then $\int_{0}^{π} f (x) sin x d x$ is equal to

The value of $\int_{- π /2}^{π /2} (\frac{1 + s i n ^{2} x}{1 + π ^{s i n x}}) d x$ is

Let $f : (0, 1) \to R$ be the functions defined as $f (x) = n$ if $x \in [\frac{1}{n + 1}, \frac{1}{n})$ where $n \in N$ . Let $g : (0, 1) \to R$ be a function such that $\int_{x^{2}}^{x} \frac{1 - t}{t} d t < g (x) < 2 x$ for all $x \in (0, 1)$ . Then $lim_{x \to 0} f (x) g (x)$

If $f$ and $g$ are continuous function on $[0, a]$ satisfying $f (x) = f (a - x)$ and $g (x) + g (a - x) = 2$ , then show that $\int_{0}^{a} f (x) g (x) d x = \int_{0}^{a} f (x) d x$

Show that $\int_{0}^{π /2} f (sin 2 x) sin x d x = 2 \int_{0}^{π /4} f (cos 2 x) cos x d x$

If $f (x) = \frac{2 - x c o s x}{2 + x c o s x}$ and $g (x) = lo g_{e} x, (x > 0)$ then the value of integral $\int_{- π /4}^{π /4} g (f (x)) d x$ is :

Animated Solution for Mathematics - Definite Integration: Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of 2∫0π/2​f(x)g(x)dx−∫0π/2​g(x)dx is ________.

Enter Numerical Value:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of π316​∫0π/2​f(x)g(x)dx is ________.

Enter Numerical Value:

Visualized Solution

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

Combine Integrals

Trigonometric Identity

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Split the Integral

First Integral: Setup

First Integral: Compute

Second Integral: Setup

Second Integral: Compute

Substitute Back into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Final Calculation

Conclusion & Discrepancy

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Art of Simplification

The Divide and Conquer Strategy

The Mastery of Integration by Parts

The Final Synthesis

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of π16S1​​ is _____.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of π248S2​​ is _____.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R and g:R→R be continuous functions. Then the value of the integral ∫−π/2π/2​[f(x)+f(−x)][g(x)−g(−x)]dx is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f and g be continuous functions on [0,a] such that f(x)=f(a−x) and g(x)+g(a−x)=4, then ∫0a​f(x)g(x)dx is equal to :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate ∫0π​2x−πxsin2xsin(2π​cosx)​dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a function satisfying f(x)+f(π−x)=π2,∀x∈R. Then ∫0π​f(x)sinxdx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ∫−π/2π/2​(1+πsinx1+sin2x​)dx is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f and g are continuous function on [0,a] satisfying f(x)=f(a−x) and g(x)+g(a−x)=2, then show that ∫0a​f(x)g(x)dx=∫0a​f(x)dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Show that ∫0π/2​f(sin2x)sinxdx=2​∫0π/4​f(cos2x)cosxdx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=2+xcosx2−xcosx​ and g(x)=loge​x,(x>0) then the value of integral ∫−π/4π/4​g(f(x))dx is :

The value of $2 \int_{0}^{π /2} f (x) g (x) d x - \int_{0}^{π /2} g (x) d x$ is ________.

The value of $\frac{16}{π ^{3}} \int_{0}^{π /2} f (x) g (x) d x$ is ________.

Define Expression $E$

Substitute $g (x)$

Substitute Back into $E$

The value of $\frac{16 S _{1}}{π}$ is _____.

The value of $\frac{48 S _{2}}{π ^{2}}$ is _____.

Let $f : R \to R$ and $g : R \to R$ be continuous functions. Then the value of the integral $\int_{- π /2}^{π /2} [f (x) + f (- x)] [g (x) - g (- x)] d x$ is

Let $f$ and $g$ be continuous functions on $[0, a]$ such that $f (x) = f (a - x)$ and $g (x) + g (a - x) = 4$ , then $\int_{0}^{a} f (x) g (x) d x$ is equal to :-

Evaluate $\int_{0}^{π} \frac{x s i n 2 x s i n ( \frac{π}{2} c o s x )}{2 x - π} d x$

Let $f (x)$ be a function satisfying $f (x) + f (π - x) = π^{2}, \forall x \in R$ . Then $\int_{0}^{π} f (x) sin x d x$ is equal to

The value of $\int_{- π /2}^{π /2} (\frac{1 + s i n ^{2} x}{1 + π ^{s i n x}}) d x$ is

If $f$ and $g$ are continuous function on $[0, a]$ satisfying $f (x) = f (a - x)$ and $g (x) + g (a - x) = 2$ , then show that $\int_{0}^{a} f (x) g (x) d x = \int_{0}^{a} f (x) d x$

Show that $\int_{0}^{π /2} f (sin 2 x) sin x d x = 2 \int_{0}^{π /4} f (cos 2 x) cos x d x$

If $f (x) = \frac{2 - x c o s x}{2 + x c o s x}$ and $g (x) = lo g_{e} x, (x > 0)$ then the value of integral $\int_{- π /4}^{π /4} g (f (x)) d x$ is :