Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
MATCH THE FOLLOWING:
(A) If $I = \int_{-1}^{1} (\alpha x^3 + \beta x + \gamma)dx$, then $I$ is
(B) Let $\alpha, \beta$ be the distinct positive roots of the equation $\tan x = 2x$, then $\gamma \int_{0}^{\alpha} (\sin \alpha x.\sin \beta x)dx$ where $\gamma \neq 0$ is
(C) If $f(x+\alpha) + f(x) = 0$, where $\alpha > 0$, then $\int_{\beta}^{\beta+2\alpha} f(x)dx$, where $\gamma \in \mathbb{N}$ is
(D) $\gamma \int_{\alpha}^{[\alpha]} [\sin x]dx$ is, where $\gamma \neq 0$, $\alpha \in [(2\beta+1)\pi, (2\beta+2)\pi]$ $n \in \mathbb{N}$, and where $[.]$ denotes the greatest integer function
Step-by-Step Solution
Key Concept: Recognize the integrand as a derivative of a product $f(x)g(x)$ and apply integration by parts with appropriate function identification.
To evaluate $\int \frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} dx$, recognize that $f(x) = \frac{x^2}{2}$ and $g(x) = \ln(x+\sqrt{1+x^2})$ are related by differentiation. Using the product rule on $f(x)g(x)$ gives $\int f(x)g(x)dx = \frac{x^3}{6}\ln(x+\sqrt{1+x^2}) - \int \frac{x^3}{6\sqrt{1+x^2}}dx$. Substitute $1+x^2 = t^2$ to evaluate the remaining integral, obtaining $\frac{x^3}{6}\ln(x+\sqrt{1+x^2}) - \frac{1}{18}(1+x^2)^{3/2} - \frac{1}{6}(1+x^2)^{1/2} + c$. Computing coefficients: $a = \frac{1}{6}$, $b = -\frac{1}{18}$, $c = -\frac{1}{6}$ yields $a+b+c = \frac{18}{5}$.
Correct Answer: [A-p, q] [B-p, q, r] [C-q, s] [D- s]