Solution & Explanation

### Related Formula

sin A + sin B = 2 sinleft(fracA+B2right) cosleft(fracA-B2right)

sin 2x = 2 sin x cos x

cos 2x = 2 cos^2 x - 1

### Core Logic Let

I = 6int_0^pi|sin 3x + sin x + sin 2x| dx

. Apply sum-to-product on

sin 3x + sin x

sin 3x + sin x = 2 sin(2x) cos(x)

So the expression becomes:

|2 sin(2x) cos x + sin 2x| = |sin 2x (2 cos x + 1)| = |2 sin x cos x (2 cos x + 1)|

Since

x in [0, pi]

sin x geq 0

. We can pull it out of the modulus.

I = 12 int_0^pi sin x |2 cos^2 x + cos x| dx

### Step 1: Coordinate Substitution Substitute

t = cos x

, then

dt = -sin x dx

. Limits: when

x = 0

t = 1

. When

x = pi

t = -1

I = 12 int_-1^1 |2t^2 + t| dt

### Step 2: Resolve Modulus Intervals The roots of

2t^2 + t = 0

are

t = 0

and

t = -1/2

. The quadratic

2t^2 + t

is negative in the interval

(-1/2, 0)

and positive elsewhere. Split the integral:

I = 12 left[ int_-1^-1/2 (2t^2 + t) dt - int_-1/2^0 (2t^2 + t) dt + int_0^1 (2t^2 + t) dt right]

Integration of modulus polynomial function diagram for Q25 - JEE Main 2026 Morning

### Step 3: Evaluate Integrals Anti-derivative:

F(t) = frac2t^33 + fract^22

F(1) = 2/3 + 1/2 = 7/6

F(0) = 0

F(-1/2) = 2(-1/8)/3 + 1/8 = -1/12 + 1/8 = 1/24

F(-1) = -2/3 + 1/2 = -1/6

Evaluate each segment: 1)

int_-1^-1/2 = F(-1/2) - F(-1) = 1/24 - (-1/6) = 1/24 + 4/24 = 5/24

-int_-1/2^0 = -(F(0) - F(-1/2)) = -(0 - 1/24) = 1/24

int_0^1 = F(1) - F(0) = 7/6 - 0 = 28/24

Sum of parts inside bracket:

frac524 + frac124 + frac2824 = frac3424 = frac1712

### Step 4: Final Output

I = 12 times frac1712 = 17

### Pattern Recognition Whenever an integral features a cascading sum of sine frequencies like

sin(kx)

, pair the highest and lowest frequencies first. The resulting common factor often matches the middle term, instantly yielding a clean polynomial substitution under

t = cos x

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Definite Integration Class 11 Maths: Trigonometric Functions

More Definite Integration Previous-Year Questions — Page 3

Q71 jee_main_2025_24_jan_morning Differentiating Under the Integral Sign

Let

f

be a differentiable function such that

2(x+2)^2f(x) - 3(x+2)^2 = 10int_0^x(t+2)f(t)dt

for

x geq 0

. Then

f(2)

is equal to ________.

Numerical Answer. Answer: 19

Solution

### Related Formula The Leibniz Integral Rule template allows direct differentiation of an integral with variable limits:

fracddxleft( int_0^x g(t) dt right) = g(x)

### Core Logic Differentiate both sides of the given functional equation with respect to

x

using the product rule:

fracddxleft[ 2(x+2)^2 f(x) - 3(x+2)^2 right] = fracddxleft[ 10int_0^x(t+2)f(t)dt right]

4(x+2)f(x) + 2(x+2)^2 f'(x) - 6(x+2) = 10(x+2)f(x)

Since

x geq 0

, the factor

(x+2)

is strictly non-zero. Divide the entire equation by

2(x+2)

2f(x) + (x+2)f'(x) - 3 = 5f(x)

(x+2)f'(x) - 3f(x) = 3

### Step 1: Solve the First-Order Differential Equation Rearrange the expression into standard linear differential equation form where

y = f(x)

fracdydx - frac3x+2y = frac3x+2

Compute the Integrating Factor (I.F.):

textI.F. = e^int -frac3x+2 dx = e^-3ln(x+2) = (x+2)^-3

Multiply through by the I.F. and integrate:

y cdot (x+2)^-3 = int frac3x+2 cdot (x+2)^-3 dx = int 3(x+2)^-4 dx

fracf(x)(x+2)^3 = 3 cdot frac(x+2)^-3-3 + C = -(x+2)^-3 + C

f(x) = -1 + C(x+2)^3

### Step 2: Apply the Boundary Condition Find the boundary condition by substituting

x = 0

into the original integral equation equation:

2(0+2)^2 f(0) - 3(0+2)^2 = 10 int_0^0 (t+2)f(t) dt

8f(0) - 12 = 0 implies f(0) = frac128 = frac32

Substitute

x = 0

into our general solution formula:

f(0) = -1 + C(0+2)^3 implies frac32 = -1 + 8C

frac52 = 8C implies C = frac516

Thus, the explicit function is:

f(x) = -1 + frac516(x+2)^3

### Step 3: Evaluate at target point x = 2 Substitute

x = 2

into the final function equation:

f(2) = -1 + frac516(2+2)^3 = -1 + frac516(64)

f(2) = -1 + 5(4) = -1 + 20 = 19

### Pattern Recognition When an equation contains a variable integral limit

int_0^x

, differentiating both sides using the Leibniz rule converts it into a standard differential equation. The initial value is found by setting

x = 0

directly in the original expression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Differential Equations

Q62 jee_main_2025_28_jan_evening Integration by Parts

Let

f:Rrightarrow R

be a twice differentiable function such that

f(2)=1

. If

F(x)=xf(x)

for all

xin R

leftint_0^2xF^prime(x)dx=6right.

and

leftint_0^2x^2F^primeprime(x)dx=40right.

, then

F^prime(2)+leftint_0^2F(x)dxright.

is equal to :

A. $11$ $11$
B. $15$ $15$
C. $6$ $6$
D. $13$ $13$

Solution

### Related Formula Integration by Parts formula:

int u cdot v \, dx = u int v \, dx - int left( u' int v \, dx right) dx

### Core Logic Given

F(x) = xf(x)

and

f(2) = 1 implies F(2) = 2f(2) = 2

. Let's apply Integration by Parts to the first given integral:

int_0^2 x F'(x) dx = 6

left[ xF(x) right]_0^2 - int_0^2 F(x) dx = 6

2F(2) - 0 - int_0^2 F(x) dx = 6

2(2) - int_0^2 F(x) dx = 6 implies int_0^2 F(x) dx = 4 - 6 = -2

### Step 1: Simplify Second Integral via Integration by Parts Now look at the second integral:

int_0^2 x^2 F''(x) dx = 40

Applying Integration by parts (taking

u = x^2

and

v = F''(x)

left[ x^2 F'(x) right]_0^2 - int_0^2 2x F'(x) dx = 40

4 F'(2) - 0 - 2 int_0^2 x F'(x) dx = 40

We already know

int_0^2 x F'(x) dx = 6

4 F'(2) - 2(6) = 40

4 F'(2) - 12 = 40 implies 4 F'(2) = 52 implies F'(2) = 13

### Step 2: Sum the Values We need to find

F'(2) + int_0^2 F(x) dx

13 + (-2) = 11

### Pattern Recognition Notice how the definition of

f(x)

is mostly a distraction to find

F(2)=2

. The problem is fundamentally testing consecutive applications of integration by parts to reduction structures. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integration

Q66 jee_main_2025_29_jan_morning Integration by Substitution

The integral

80int_0^fracpi4left(fracsintheta + costheta9 + 16sin 2thetaright)mathrmdtheta

is equal to:

A. $3log_e4$ $3\log_{e}4$
B. $6log_mathrme4$ $6\log_{\mathrm{e}4}$
C. $4log_mathrme3$ $4\log_{\mathrm{e}3}$
D. $2log_e3$ $2\log_{e}3$

Solution

### Related Formula

int fracdta^2 - b^2 t^2 = frac12a ln left| fraca+bta-bt right|

sin 2theta = 1 - (sintheta - costheta)^2

### Core Logic Let

sintheta - costheta = t

. Then

(costheta + sintheta)dtheta = dt

. Transform limits: When

theta = 0 implies t = 0 - 1 = -1

When

theta = fracpi4 implies t = frac1sqrt2 - frac1sqrt2 = 0

### Step 1: Perform the algebraic substitution Express the denominator base:

9 + 16sin 2theta = 9 + 16[1 - t^2] = 25 - 16t^2

The integral transforms to:

I = 80 int_-1^0 fracdt25 - 16t^2 = frac8016 int_-1^0 fracdtleft(frac54right)^2 - t^2

### Step 2: Execute Integral Calculation

I = 5 left[ frac12left(frac54right) ln left| fracfrac54 + tfrac54 - t right| right]_-1^0

I = 2 left[ ln(1) - lnleft( frac1/49/4 right) right] = 2 left[ 0 - lnleft(frac19right) right] = 2ln(9) = 4ln(3)

### Pattern Recognition Whenever

(sintheta + costheta)

sits inside the numerator, instantly choose

t = sintheta - costheta

as your core linear substitution engine to clean denominators. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals

Q71 jee_main_2025_29_jan_morning Functional Equations with Integrals

Let

f: (0, infty) to mathbfR

be a twice differentiable function. If for some

a neq 0

int_0^1 f(lambda x) \, mathrmdlambda = a f(x)

f(1) = 1

and

f(16) = frac18

, then

16 - f'(frac116)

is equal to

Numerical Answer. Answer: 112

Solution

### Related Formula

textLeibniz Integral Rule for differentiation: fracddxint_0^x f(t) \, dt = f(x)

### Core Logic Perform variable substitution inside the integral: let

lambda x = t implies dlambda = frac1x dt

. When

lambda = 0 implies t = 0

; when

lambda = 1 implies t = x

. The equation transforms to:

frac1x int_0^x f(t) \, dt = a f(x) implies int_0^x f(t) \, dt = a x f(x)

### Step 1: Differentiate with respect to x Using Leibniz rule and product rule:

f(x) = a [x f'(x) + f(x)]

(1 - a)f(x) = a x f'(x) implies fracf'(x)f(x) = frac1-aa frac1x

Integrating both sides yields:

ln f(x) = left(frac1-aaright)ln x + c implies f(x) = C x^frac1-aa

### Step 2: Calculate Constants using boundaries Given

f(1) = 1 implies C = 1

. Given

f(16) = frac18 implies frac18 = (16)^frac1-aa implies 2^-3 = (2^4)^frac1-aa

-3 = frac4(1-a)a implies -3a = 4 - 4a implies a = 4

Therefore, power exponent

= frac1-44 = -frac34 implies f(x) = x^-frac34

. ### Step 3: Evaluate target derivative value Find the derivative:

f'(x) = -frac34 x^-frac74

Substitute

x = frac116

f'left(frac116right) = -frac34 left(2^-4right)^-frac74 = -frac34 left(2^7right) = -frac34 times 128 = -96

Final requested computation calculation:

16 - f'left(frac116right) = 16 - (-96) = 112

### Pattern Recognition Scaling inputs inside functional definite integrals tracks closely to homogenous Euler equation properties. Converting integrations quickly to local power functions reduces processing parameters. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Differential Equations

Q2 jee_main_2024_01_february_morning Properties of Definite Integrals

The value of the integral

int_0^fracpi4fracx \, dxsin^4(2x)+cos^4(2x)

equals:

A. $fracsqrt2pi^28$ $\frac{\sqrt{2}\pi^{2}}{8}$
B. $fracsqrt2pi^216$ $\frac{\sqrt{2}\pi^{2}}{16}$
C. $fracsqrt2pi^232$ $\frac{\sqrt{2}\pi^{2}}{32}$
D. $fracsqrt2pi^264$ $\frac{\sqrt{2}\pi^{2}}{64}$

Solution

### Related Formula King's Property of Definite Integrals:

int_a^b f(x) \, dx = int_a^b f(a+b-x) \, dx

### Core Logic Let the given integral be

I

I = int_0^fracpi4fracx \, dxsin^4(2x)+cos^4(2x)

Substitute

2x = t implies 2dx = dt implies dx = frac12dt

. When

x = 0 implies t = 0

, and when

x = fracpi4 implies t = fracpi2

I = frac14 int_0^fracpi2 fract \, dtsin^4t + cos^4t quad implies (1)

### Step 1: Apply King's Property Applying the property

int_0^a f(t) \, dt = int_0^a f(a-t) \, dt

I = frac14 int_0^fracpi2 fracleft(fracpi2 - tright) dtsin^4left(fracpi2-tright) + cos^4left(fracpi2-tright)

I = frac14 int_0^fracpi2 fracleft(fracpi2 - tright) dtcos^4t + sin^4t quad implies (2)

Adding equations (1) and (2):

2I = frac14 int_0^fracpi2 fracfracpi2 dtsin^4t + cos^4t

2I = fracpi8 int_0^fracpi2 fracdtsin^4t + cos^4t

2I = fracpi8 int_0^fracpi2 fracsec^4t \, dttan^4t + 1

2I = fracpi8 int_0^fracpi2 frac(1 + tan^2t)sec^2t \, dttan^4t + 1

### Step 2: Substitution and Algebraic Limits Let

tan t = y implies sec^2t \, dt = dy

. When

t = 0 implies y = 0

, and when

t = fracpi2 implies y = infty

2I = fracpi8 int_0^infty frac(1 + y^2) \, dy1 + y^4

I = fracpi16 int_0^infty frac1 + frac1y^2y^2 + frac1y^2 \, dy

Now put

y - frac1y = p implies left(1 + frac1y^2right) dy = dp

. When

y to 0^+ implies p to -infty

, and when

y to infty implies p to infty

. Also,

y^2 + frac1y^2 = p^2 + 2 = p^2 + (sqrt2)^2

I = fracpi16 int_-infty^infty fracdpp^2 + (sqrt2)^2

I = fracpi16sqrt2 left[ tan^-1left(fracpsqrt2right) right]_-infty^infty

I = fracpi16sqrt2 left( fracpi2 - left(-fracpi2right) right) = fracpi^216sqrt2 = fracsqrt2pi^232

### Pattern Recognition Sees: Integrand containing

x

in the numerator and symmetric trigonometric functions in the denominator. Shortcut: The elimination of

x

using King's property is standard. For integrals containing

frac1+y^21+y^4

, dividing by

y^2

transforms the denominator into a perfect square form

left(y-frac1yright)^2+2

, making substitution trivial. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 11 Mathematics: Trigonometric Identities

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