Given, I=∫cos(lnx)dx
Let, lnx=t⇒x=et⇒dx=etdt
∴I=∫etcostdt
∫etcostdt=21∫et((cost+sint)+(cost−sint))dt
∫etcostdt=21et(cost+sint)+C, where C is the constant of integration.
=2x(cos(lnx)+sin(lnx))+C.
JEE Main 2019 — Mathematics Calculus
The integral ∫cos(lnx)dx, is equal to
Held on 12 Jan 2019 · Verified 6 Jul 2026.
2x(cos(lnx)−sin(lnx))+C
x(cos(lnx)−sin(lnx))+C
x(cos(lnx)+sin(lnx))+C
2x(cos(lnx)+sin(lnx))+C
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
The value of $\int_{-\pi / 6}^{\pi / 6}\left(\frac{\pi+4 x^{11}}{1-\sin (|x|+\pi / 6)}\right) d x$ is equal to:
The product of all possible values of $\alpha$, for which $\displaystyle\lim_{x \to 0}\left(\dfrac{1 - \cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)}{\sin^2((\alpha+1)x)}\right) = 2$, is:
The value of the integral $\displaystyle\int_0^\infty \dfrac{\log_e(x)}{x^2 + 4}\,dx$ is:
Let $f$ be a differentiable function satisfying $f(x)=1-2 x+\int_{0}^{x} \mathrm{e}^{(x-t)} f(t) \mathrm{dt}, x \in \mathbf{R}$ and let $\mathrm{g}(x)=\int_{0}^{x}(f(\mathrm{t})+2)^{15}(\mathrm{t}-4)^{6}(\mathrm{t}+12)^{17} \mathrm{dt}, x \in \mathbf{R}$. If p and q are respectively the points of local minima and local maxima of g, then the value of $|\mathrm{p}+\mathrm{q}|$ is equal to $\_\_\_\_$.
Let the area of the region bounded by the curve $y=\max \{\sin x, \cos x\}$, lines $x=0, x=\frac{3 \pi}{2}$, and the $x$-axis be A. Then, $\mathrm{A}+\mathrm{A}^{2}$ is equal to $\_\_\_\_$.
Work through every JEE Main Calculus PYQ, year by year.