Let the domain of the function f(x) = log_2log_4log_6(3 + 4x - x^2)$f(x) = \log_{2}\log_{4}\log_{6}(3 + 4x - x^{2})$ be (a, b)$(a, b)$[cite: 663]. If int_0^b-a[x^2]dx = p - sqrtq - sqrtr$\int_{0}^{b-a}[x^{2}]dx = p - \sqrt{q} - \sqrt{r}$ [cite: 664], where p, q, r in mathbbN$p, q, r \in \mathbb{N}$ [cite: 664] and gcd(p, q, r) = 1$\gcd(p, q, r) = 1$ [cite: 668], and [cdot]$[\cdot]$ represents the greatest integer function [cite: 668], then p + q + r$p + q + r$ is equal to[cite: 668]:
A.10
B.8
C.11
D.9
Solution & Explanation
### Related Formula
Domain of log chain iterations: For log_2log_4(M) > 0$\log_2\log_4(M) > 0$, we require log_4(M) > 1 implies M > 4$\log_4(M) > 1 \implies M > 4$.
### Core Logic
Trace internal arguments outward sequentially [cite: 1393, 1394]:
log_4log_6(3 + 4x - x^2) > 0 implies log_6(3 + 4x - x^2) > 1$$\log_{4}\log_{6}(3 + 4x - x^2) > 0 \implies \log_{6}(3 + 4x - x^2) > 1$$ [cite: 1393, 1394]
3 + 4x - x^2 > 6^1 implies x^2 - 4x + 3 < 0$$3 + 4x - x^2 > 6^1 \implies x^2 - 4x + 3 < 0$$ [cite: 1395, 1396]
(x-1)(x-3) < 0 implies x in (1, 3)$$(x-1)(x-3) < 0 \implies x \in (1, 3)$$ [cite: 1397, 1398]
Thus, determine limits [cite: 1399]:
a = 1, quad b = 3 implies b - a = 2$$a = 1, \quad b = 3 \implies b - a = 2$$ [cite: 1399]
### Step 1: Setting up the greatest integer function integration
We need to evaluate int_0^2 [x^2] \, mathrmdx$\int_0^2 [x^2] \, \mathrm{d}x$[cite: 1400]. Identify step boundary switch locations inside range [0, 2]$[0, 2]$ [cite: 1400]:
- For x in [0, 1): [x^2] = 0$x \in [0, 1): [x^2] = 0$
- For x in [1, sqrt2): [x^2] = 1$x \in [1, \sqrt{2}): [x^2] = 1$
- For x in [sqrt2, sqrt3): [x^2] = 2$x \in [\sqrt{2}, \sqrt{3}): [x^2] = 2$
- For x in [sqrt3, 2): [x^2] = 3$x \in [\sqrt{3}, 2): [x^2] = 3$
Set up separate boundary component integrations [cite: 1400]:
int_0^2 [x^2] \, mathrmdx = int_0^1 0 \, mathrmdx + int_1^sqrt2 1 \, mathrmdx + int_sqrt2^sqrt3 2 \, mathrmdx + int_sqrt3^2 3 \, mathrmdx$$\int_0^2 [x^2] \, \mathrm{d}x = \int_0^1 0 \, \mathrm{d}x + \int_1^{\sqrt{2}} 1 \, \mathrm{d}x + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, \mathrm{d}x + \int_{\sqrt{3}}^2 3 \, \mathrm{d}x$$ [cite: 1400]
= 0 + (sqrt2 - 1) + 2(sqrt3 - sqrt2) + 3(2 - sqrt3)$$= 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3})$$ [cite: 1400]
= sqrt2 - 1 + 2sqrt3 - 2sqrt2 + 6 - 3sqrt3 = 5 - sqrt2 - sqrt3$$= \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} = 5 - \sqrt{2} - \sqrt{3}$$ [cite: 1400]
### Step 2: Matching coefficients
Compare values with requested answer template shape [cite: 1400]:
5 - sqrt2 - sqrt3 = p - sqrtq - sqrtr$$5 - \sqrt{2} - \sqrt{3} = p - \sqrt{q} - \sqrt{r}$$ [cite: 1400]
p = 5, quad q = 2, quad r = 3$$p = 5, \quad q = 2, \quad r = 3$$ [cite: 1400]
textFinal Sum = p + q + r = 5 + 2 + 3 = 10$$\text{Final Sum} = p + q + r = 5 + 2 + 3 = 10$$ [cite: 1400]
### Pattern Recognition
Integrals over greatest integer configurations change value exactly where the inner expression tracks through integer milestones. Mapping boundaries accurately resolves calculations smoothly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integrals
Keywords:#domain nested log step function definite integral#JEE Main 2025 Morning Q67#Integrals JEE Main 2025#Definite Integral of Greatest Integer Function
More Integrals 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 Integrals Questions — jee_main_2025_03_april_morning
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