The area of the region enclosed by the parabola (y - 2)^2 = x - 1$(y - 2)^2 = x - 1$ , the line x - 2y + 4 = 0$x - 2y + 4 = 0$ and the positive coordinate axes is
Numerical Answer Type:
Enter a numerical valueAnswer: 5 to 5+4 marks
Solution & Explanation
### Related Formula
textArea enclosed integrating w.r.t y-axis: int (x_textright - x_textleft) dy$\text{Area enclosed integrating w.r.t y-axis: } \int (x_{\text{right}} - x_{\text{left}}) dy$
### Core Logic
Find intersection points between the parabola x = (y-2)^2 + 1$x = (y-2)^2 + 1$ and the line x = 2y - 4$x = 2y - 4$.
Set them equal:
(y-2)^2 + 1 = 2y - 4$(y-2)^2 + 1 = 2y - 4$y^2 - 4y + 4 + 1 = 2y - 4$y^2 - 4y + 4 + 1 = 2y - 4$y^2 - 6y + 9 = 0 Rightarrow (y-3)^2 = 0$y^2 - 6y + 9 = 0 \Rightarrow (y-3)^2 = 0$
The line is tangent to the parabola at y = 3$y = 3$. At y = 3$y = 3$, x = 2(3) - 4 = 2$x = 2(3) - 4 = 2$.
We need the area enclosed by the parabola, the line, and the *positive coordinate axes*.
Let's trace the boundary intercepts.
The line x = 2y - 4$x = 2y - 4$ cuts the y-axis (x=0$x=0$) at y=2$y=2$.
Parabola vertex is at (1, 2)$(1, 2)$. Parabola cuts y-axis (x=0$x=0$) at (y-2)^2 = -1$(y-2)^2 = -1$ (No real y-intercept).
So the curve (y-2)^2 = x-1$(y-2)^2 = x-1$ only exists for x ge 1$x \ge 1$.
### Step 1: Setting up the Integral w.r.t y
Area Under Curve diagram for Q29 - JEE Main 2024 Evening
Because the area is bound by the positive coordinate axes, we track integration strictly in the first quadrant.
The required area is the area between the y-axis, the x-axis, the line, and the curve. Wait, the problem says "area enclosed by the parabola... line... and positive coordinate axes".
Let's integrate with respect to y$y$.
The boundary curves are x = (y-2)^2 + 1$x = (y-2)^2 + 1$ (which forms the rightmost boundary) and the axes.
The area bounded by the curve with y-axis is from y=0$y=0$ to y=3$y=3$ minus the triangular region cut by the line below y=2$y=2$.
The line intersects the x-axis (y=0$y=0$) at x=-4$x=-4$ and y-axis (x=0$x=0$) at y=2$y=2$.
The positive coordinate axes enforce boundaries at x ge 0, y ge 0$x \ge 0, y \ge 0$.
So the exact area is the integral of the parabola's x$x$-value from y=0$y=0$ to y=3$y=3$, minus the area of the small triangle outside the region bounded by the line x=2y-4$x=2y-4$ and the axes in the positive quadrant.
The area under the parabola (to the left towards the y-axis) from y=0$y=0$ to y=3$y=3$ is int_0^3 x_textparabola dy$\int_0^3 x_{\text{parabola}} dy$.
The line bounds the region on the left from y=2$y=2$ to y=3$y=3$. Between y=0$y=0$ and y=2$y=2$, the area goes fully to the y-axis.
So, area = int_0^3 x_textparabola dy - textArea of Delta text bounded by line and y-axis$\int_0^3 x_{\text{parabola}} dy - \text{Area of } \Delta \text{ bounded by line and y-axis}$.
The triangle bounded by x=2y-4, x=0, y=2, y=3$x=2y-4, x=0, y=2, y=3$ is: base is x=2(3)-4 = 2$x=2(3)-4 = 2$ at y=3$y=3$, height is 1$1$ (from y=2$y=2$ to y=3$y=3$). Triangle area = frac12 times 2 times 1 = 1$\frac{1}{2} \times 2 \times 1 = 1$.
Hence, Enclosed Area = int_0^3 ((y-2)^2 + 1) dy - 1$\int_{0}^{3} ((y-2)^2 + 1) dy - 1$.
### Step 2: Evaluating the Integral
textArea = int_0^3 (y^2 - 4y + 5) dy - 1$\text{Area} = \int_{0}^{3} (y^2 - 4y + 5) dy - 1$= left[ fracy^33 - 2y^2 + 5y right]_0^3 - 1$= \left[ \frac{y^3}{3} - 2y^2 + 5y \right]_0^3 - 1$= left( frac273 - 2(9) + 5(3) right) - 0 - 1$= \left( \frac{27}{3} - 2(9) + 5(3) \right) - 0 - 1$= (9 - 18 + 15) - 1 = 6 - 1 = 5$= (9 - 18 + 15) - 1 = 6 - 1 = 5$
### Pattern Recognition
Instead of evaluating horizontal strips with a broken left-bound, evaluate the total area mapped to the y-axis int_0^3 x dy$\int_0^3 x dy$ and simply subtract the basic geometric triangle removed by the straight line.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Integral Calculus
Keywords:#Area bounded by parabola and line#JEE Main 2024 Evening Q29#Integral Calculus JEE Main 2024#Area Under Curve JEE Main 2024
More Integral Calculus Previous-Year Questions — Page 3
Q72jee_main_2025_07_april_eveningIntegration by Substitution
If int left(frac1x +frac1x^3right)left(sqrt[2]3x^-24 + x^-26right)dx = - frac alpha3 (alpha + 1) left(3 x ^ beta + x ^ gammaright) ^ frac alpha + 1alpha + C$\int \left(\frac{1}{x} +\frac{1}{x^3}\right)\left(\sqrt[2]{3x^{-24} + x^{-26}}\right)dx = - \frac {\alpha}{3 (\alpha + 1)} \left(3 x ^ {\beta} + x ^ {\gamma}\right) ^ {\frac {\alpha + 1}{\alpha}} + C$, mathbfx > 0, (alpha, beta, gamma in mathbfZ)$\mathbf{x} > 0, (\alpha, \beta, \gamma \in \mathbf{Z})$, where mathbfC$\mathbf{C}$ is the constant of integration, then alpha + beta + gamma$\alpha + \beta + \gamma$ is equal to
Numerical Answer.Answer: 19 to 19
Solution
### Related Formula
Standard power integration formula rule is:
int u^n \, du = fracu^n+1n+1 + C$\int u^n \, du = \frac{u^{n+1}}{n+1} + C$
### Core Logic
Rewrite the integral by adjusting powers inside the radical container:
I = int left(frac1x^2 + frac1x^4right) left(frac3x + frac1x^3right)^frac123 \, dx$I = \int \left(\frac{1}{x^2} + \frac{1}{x^4}\right) \left(\frac{3}{x} + \frac{1}{x^3}\right)^{\frac{1}{23}} \, dx$
Let t = frac3x + frac1x^3 implies dt = -3left(frac1x^2 + frac1x^4right)\,dx$t = \frac{3}{x} + \frac{1}{x^3} \implies dt = -3\left(\frac{1}{x^2} + \frac{1}{x^4}\right)\,dx$.
### Step 1: Integrate
Substituting t$t$ into the equation:
int fract^1/23 \, dt-3 = -frac13 cdot fract^24/2324/23 = -frac233(24) t^frac2423$\int \frac{t^{1/23} \, dt}{-3} = -\frac{1}{3} \cdot \frac{t^{24/23}}{24/23} = -\frac{23}{3(24)} t^{\frac{24}{23}}$
Comparing parameters directly yields:
alpha = 23, quad beta = -1, quad gamma = -3$\alpha = 23, \quad \beta = -1, \quad \gamma = -3$alpha + beta + gamma = 23 - 1 - 3 = 19$\alpha + \beta + \gamma = 23 - 1 - 3 = 19$
### Pattern Recognition
Pull out high powers from the root factor to seamlessly reveal a matching f'(x)$f'(x)$ substitution pair on the outside.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integral Calculus
Q75jee_main_2025_24_jan_eveningIndefinite Integration of Algebraic Functions
If intfrac2x^2+5x+9sqrtx^2+x+1dx=xsqrtx^2+x+1+alphasqrtx^2+x+1+betalog_eleft|x+frac12+sqrtx^2+x+1right|+C$\int\frac{2x^{2}+5x+9}{\sqrt{x^{2}+x+1}}dx=x\sqrt{x^{2}+x+1}+\alpha\sqrt{x^{2}+x+1}+\beta\log_{e}\left|x+\frac{1}{2}+\sqrt{x^{2}+x+1}\right|+C$ where C is the constant of integration, then alpha+2beta$\alpha+2\beta$ is equal to \_\_\_\_. [cite: 3419, 3420]
Numerical Answer.Answer: 16
Solution
### Related Formula
Standard integration templates for quadratic forms:
int sqrtt^2 + k^2 dt = fract2sqrtt^2+k^2 + frack^22lnleft|t + sqrtt^2+k^2right|$\int \sqrt{t^2 + k^2} dt = \frac{t}{2}\sqrt{t^2+k^2} + \frac{k^2}{2}\ln\left|t + \sqrt{t^2+k^2}\right|$int frac1sqrtt^2 + k^2 dt = lnleft|t + sqrtt^2+k^2right|$\int \frac{1}{\sqrt{t^2 + k^2}} dt = \ln\left|t + \sqrt{t^2+k^2}\right|$
### Core Logic
Decompose the numerator using polynomial differentiation components :
2x^2 + 5x + 9 = A(x^2 + x + 1) + B(2x + 1) + C$2x^2 + 5x + 9 = A(x^2 + x + 1) + B(2x + 1) + C$
Equating coefficients dynamically yields :
- For x^2$x^2$: A = 2$A = 2$ .
- For x$x$: A + 2B = 5 Rightarrow 2 + 2B = 5 Rightarrow B = frac32$A + 2B = 5 \Rightarrow 2 + 2B = 5 \Rightarrow B = \frac{3}{2}$ .
- Constant: A + B + C = 9 Rightarrow 2 + frac32 + C = 9 Rightarrow C = frac112$A + B + C = 9 \Rightarrow 2 + \frac{3}{2} + C = 9 \Rightarrow C = \frac{11}{2}$ .
Rewrite the integrand into three parts :
2int sqrtx^2+x+1 dx + frac32int frac2x+1sqrtx^2+x+1 dx + frac112int frac1sqrtx^2+x+1 dx$2\int \sqrt{x^2+x+1} dx + \frac{3}{2}\int \frac{2x+1}{\sqrt{x^2+x+1}} dx + \frac{11}{2}\int \frac{1}{\sqrt{x^2+x+1}} dx$
### Step 1: Complete Quadratics & Integrate
Format using completing the square technique: x^2 + x + 1 = left(x + frac12right)^2 + left(fracsqrt32right)^2$x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$.
1. First Integral evaluation:
2 left[ fracleft(x+frac12right)2sqrtx^2+x+1 + frac38lnleft|x+frac12+sqrtx^2+x+1right| right]$2 \left[ \frac{\left(x+\frac{1}{2}\right)}{2}\sqrt{x^2+x+1} + \frac{3}{8}\ln\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right| \right]$= left(x+frac12right)sqrtx^2+x+1 + frac34lnleft|x+frac12+sqrtx^2+x+1right|$= \left(x+\frac{1}{2}\right)\sqrt{x^2+x+1} + \frac{3}{4}\ln\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|$
2. Second Integral evaluation :
frac32 cdot 2sqrtx^2+x+1 = 3sqrtx^2+x+1$\frac{3}{2} \cdot 2\sqrt{x^2+x+1} = 3\sqrt{x^2+x+1}$
3. Third Integral evaluation :
frac112lnleft|x+frac12+sqrtx^2+x+1right|$\frac{11}{2}\ln\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|$
### Step 2: Collect Like Terms
Gather and combine matching factor parameters:
textTotal = left(x + frac12 + 3right)sqrtx^2+x+1 + left(frac34 + frac112right)lnleft|x+frac12+sqrtx^2+x+1right|$\text{Total} = \left(x + \frac{1}{2} + 3\right)\sqrt{x^2+x+1} + \left(\frac{3}{4} + \frac{11}{2}\right)\ln\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|$textTotal = left(x + frac72right)sqrtx^2+x+1 + frac254lnleft|x+frac12+sqrtx^2+x+1right|$\text{Total} = \left(x + \frac{7}{2}\right)\sqrt{x^2+x+1} + \frac{25}{4}\ln\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|$= xsqrtx^2+x+1 + frac72sqrtx^2+x+1 + frac254lnleft|x+frac12+sqrtx^2+x+1right|$= x\sqrt{x^2+x+1} + \frac{7}{2}\sqrt{x^2+x+1} + \frac{25}{4}\ln\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|$
### Step 3: Extract Coefficients
Compare directly against the given expression variables :
alpha = frac72, quad beta = frac254$\alpha = \frac{7}{2}, \quad \beta = \frac{25}{4}$alpha + 2beta = frac72 + 2left(frac254right) = frac72 + frac252 = frac322 = 16$\alpha + 2\beta = \frac{7}{2} + 2\left(\frac{25}{4}\right) = \frac{7}{2} + \frac{25}{2} = \frac{32}{2} = 16$
### Pattern Recognition
When dividing large numerators containing x^2$x^2$ elements over quadratic square roots, using matching coefficient expansion rules prevents lengthy substitution errors completely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integrals
Q57jee_main_2025_28_jan_eveningLeibnitz Rule and Definite Integral
Let f be a real valued continuous function defined on the positive real axis such that g(x)=int_0^xtf(t)dt$g(x)=\int_{0}^{x}tf(t)dt$. If g(x^3)=x^6+x^7$g(x^{3})=x^{6}+x^{7}$ then value of sum_r=1^15f(r^3)$\sum_{r=1}^{15}f(r^{3})$ is:
A.320$320$
B.340$340$
C.270$270$
D.310$310$
Solution
### Related Formula
Newton-Leibnitz Theorem for differentiation under integral sign:
fracddx left( int_0^x tf(t) dt right) = xf(x)$\frac{d}{dx} \left( \int_{0}^{x} tf(t) dt \right) = xf(x)$
### Core Logic
Given:
g(x) = int_0^x tf(t) dt$g(x) = \int_{0}^{x} tf(t) dt$
Differentiating both sides with respect to x$x$:
g'(x) = xf(x) implies f(x) = fracg'(x)x$g'(x) = xf(x) \implies f(x) = \frac{g'(x)}{x}$
We are given g(x^3) = x^6 + x^7$g(x^3) = x^6 + x^7$. Let y = x^3 implies x = y^1/3$y = x^3 \implies x = y^{1/3}$.
Substituting this into the expression for g$g$:
g(y) = (y^1/3)^6 + (y^1/3)^7 = y^2 + y^7/3$g(y) = (y^{1/3})^6 + (y^{1/3})^7 = y^2 + y^{7/3}$
Thus, replacing y$y$ back with x$x$:
g(x) = x^2 + x^7/3$g(x) = x^2 + x^{7/3}$
### Step 1: Differentiate g(x) to find f(x)
g'(x) = 2x + frac73x^4/3$g'(x) = 2x + \frac{7}{3}x^{4/3}$
Now find f(x)$f(x)$:
f(x) = fracg'(x)x = frac2x + frac73x^4/3x = 2 + frac73x^1/3$f(x) = \frac{g'(x)}{x} = \frac{2x + \frac{7}{3}x^{4/3}}{x} = 2 + \frac{7}{3}x^{1/3}$
### Step 2: Evaluate the Summation
We need to find sum_r=1^15 f(r^3)$\sum_{r=1}^{15} f(r^3)$:
f(r^3) = 2 + frac73(r^3)^1/3 = 2 + frac73r$f(r^3) = 2 + \frac{7}{3}(r^3)^{1/3} = 2 + \frac{7}{3}r$
Now, compute the summation from r=1$r=1$ to 15$15$:
sum_r=1^15 f(r^3) = sum_r=1^15 left( 2 + frac73r right) = sum_r=1^15 2 + frac73sum_r=1^15 r$\sum_{r=1}^{15} f(r^3) = \sum_{r=1}^{15} \left( 2 + \frac{7}{3}r \right) = \sum_{r=1}^{15} 2 + \frac{7}{3}\sum_{r=1}^{15} r$(2 times 15) + frac73 times frac15 times 162$(2 \times 15) + \frac{7}{3} \times \frac{15 \times 16}{2}$30 + frac73 times 120 = 30 + 7 times 40 = 30 + 280 = 310$30 + \frac{7}{3} \times 120 = 30 + 7 \times 40 = 30 + 280 = 310$
### Pattern Recognition
Converting g(x^3)$g(x^3)$ directly into a function of variable y=x^3$y=x^3$ prevents multi-layer chain rule complications when applying differentiation immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Class 12 Mathematics: Definite Integration
Q61jee_main_2025_28_jan_eveningIntegration by Substitution
If f(x)=leftintfrac1x^1/4(1+x^1/4)dxright.$f(x)=\left\int\frac{1}{x^{1/4}(1+x^{1/4})}dx\right.$, f(0)=-6,$f(0)=-6,$ then f(1)$f(1)$ is equal to:
A.log_e2+2$\log_{e}2+2$
B.4(log_e2-2)$4(\log_{e}2-2)$
C.2-log_e2$2-\log_{e}2$
D.4(log_e2+2)$4(\log_{e}2+2)$
Solution
### Related Formula
Standard substitution method and logarithmic integral rule:
int frac1t+1 dt = ln|t+1| + C$\int \frac{1}{t+1} dt = \ln|t+1| + C$
### Core Logic
Let x = t^4 implies dx = 4t^3 dt$x = t^4 \implies dx = 4t^3 dt$.
When substituting into the integral:
f(x) = int frac4t^3t(1+t) dt = 4 int fract^21+t dt$f(x) = \int \frac{4t^3}{t(1+t)} dt = 4 \int \frac{t^2}{1+t} dt$
### Step 1: Simplify the Integral
Rewrite the numerator t^2$t^2$ as (t^2 - 1) + 1$(t^2 - 1) + 1$:
4 int frac(t^2 - 1) + 11+t dt = 4 int left( frac(t-1)(t+1)1+t + frac11+t right) dt$4 \int \frac{(t^2 - 1) + 1}{1+t} dt = 4 \int \left( \frac{(t-1)(t+1)}{1+t} + \frac{1}{1+t} \right) dt$4 int (t - 1) dt + 4 int frac1t+1 dt$4 \int (t - 1) dt + 4 \int \frac{1}{t+1} dt$4 left[ frac(t-1)^22 right] + 4 ln|t+1| + C = 2(t-1)^2 + 4 ln|t+1| + C$4 \left[ \frac{(t-1)^2}{2} \right] + 4 \ln|t+1| + C = 2(t-1)^2 + 4 \ln|t+1| + C$
Substitute back t = x^1/4$t = x^{1/4}$:
f(x) = 2(x^1/4 - 1)^2 + 4 ln(1 + x^1/4) + C$f(x) = 2(x^{1/4} - 1)^2 + 4 \ln(1 + x^{1/4}) + C$
### Step 2: Solve for Constant C and Find f(1)
Given f(0) = -6$f(0) = -6$:
-6 = 2(0 - 1)^2 + 4 ln(1 + 0) + C$-6 = 2(0 - 1)^2 + 4 \ln(1 + 0) + C$-6 = 2 + 0 + C implies C = -8$-6 = 2 + 0 + C \implies C = -8$
Now find f(1)$f(1)$:
f(1) = 2(1^1/4 - 1)^2 + 4 ln(1 + 1^1/4) - 8$f(1) = 2(1^{1/4} - 1)^2 + 4 \ln(1 + 1^{1/4}) - 8$f(1) = 2(0) + 4 ln(2) - 8 = 4 ln 2 - 8 = 4(ln 2 - 2)$f(1) = 2(0) + 4 \ln(2) - 8 = 4 \ln 2 - 8 = 4(\ln 2 - 2)$
### Pattern Recognition
By adding and subtracting terms in the numerator (t^2-1+1$t^2-1+1$), we can quickly bypass long division for polynomials and directly integrate using standard forms.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Indefinite Integration
If int fracsin^frac32 x + cos^frac32 xsqrtsin^3 x cos^3 x sin (x - theta) d x = A sqrtcos theta tan x - sin theta + B sqrtcos theta - sin theta cot x + C$\int \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sqrt{\sin^3 x \cos^3 x \sin (x - \theta)}} d x = A \sqrt{\cos \theta \tan x - \sin \theta} + B \sqrt{\cos \theta - \sin \theta \cot x} + C$, where C$C$ is the integration constant, then AB$AB$ is equal to
### Related Formula
sin(x - theta) = sin x cos theta - cos x sin theta$\sin(x - \theta) = \sin x \cos \theta - \cos x \sin \theta$
### Core Logic
Let us partition the given integral I$I$ into two clean parts I_1$I_1$ and I_2$I_2$:
I = int fracsin^3/2 xsqrtsin^3 x cos^3 x (sin x cos theta - cos x sin theta) \, dx + int fraccos^3/2 xsqrtsin^3 x cos^3 x (sin x cos theta - cos x sin theta) \, dx$I = \int \frac{\sin^{3/2} x}{\sqrt{\sin^3 x \cos^3 x (\sin x \cos \theta - \cos x \sin \theta)}} \, dx + \int \frac{\cos^{3/2} x}{\sqrt{\sin^3 x \cos^3 x (\sin x \cos \theta - \cos x \sin \theta)}} \, dx$
Factoring out appropriate powers of sin x$\sin x$ and cos x$\cos x$ from inside the roots:
I_1 = int fracsec^2 xsqrttan x cos theta - sin theta \, dx$I_1 = \int \frac{\sec^2 x}{\sqrt{\tan x \cos \theta - \sin \theta}} \, dx$I_2 = int fraccsc^2 xsqrtcos theta - cot x sin theta \, dx$I_2 = \int \frac{\csc^2 x}{\sqrt{\cos \theta - \cot x \sin \theta}} \, dx$
### Step 1: Evaluating the Integrals
For I_1$I_1$, substitute tan x cos theta - sin theta = t^2 implies sec^2 x \, dx = frac2t \, dtcos theta$\tan x \cos \theta - \sin \theta = t^2 \implies \sec^2 x \, dx = \frac{2t \, dt}{\cos \theta}$:
I_1 = int frac2t \, dtt cos theta = frac2tcos theta = 2sec theta sqrttan x cos theta - sin theta$I_1 = \int \frac{2t \, dt}{t \cos \theta} = \frac{2t}{\cos \theta} = 2\sec \theta \sqrt{\tan x \cos \theta - \sin \theta}$
For I_2$I_2$, substitute cos theta - cot x sin theta = z^2 implies csc^2 x \, dx = frac2z \, dzsin theta$\cos \theta - \cot x \sin \theta = z^2 \implies \csc^2 x \, dx = \frac{2z \, dz}{\sin \theta}$:
I_2 = int frac2z \, dzz sin theta = frac2zsin theta = 2csc theta sqrtcos theta - cot x sin theta$I_2 = \int \frac{2z \, dz}{z \sin \theta} = \frac{2z}{\sin \theta} = 2\csc \theta \sqrt{\cos \theta - \cot x \sin \theta}$
### Step 2: Combining Coefficients
Comparing with the given expression, we find the coefficients:
A = 2sec theta, quad B = 2csc theta$A = 2\sec \theta, \quad B = 2\csc \theta$
Multiplying them together:
AB = 4sec theta csc theta = frac4sin theta cos theta = frac82sin theta cos theta = 8csc(2theta)$AB = 4\sec \theta \csc \theta = \frac{4}{\sin \theta \cos \theta} = \frac{8}{2\sin \theta \cos \theta} = 8\csc(2\theta)$
### Pattern Recognition
When integrating roots of trigonometric functions involving (x - theta)$(x - \theta)$, try dividing/multiplying fields by cos^n x$\cos^n x$ or sin^n x$\sin^n x$ to force tan x / sec^2 x$\tan x / \sec^2 x$ templates.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integrals
More Integral Calculus Questions — jee_main_2024_30_january_evening
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