Question:
if \(y=\int\frac{e^{\tan x}dx}{\cos^2x(1+e^{2\tan x})}\) and y(0)=6 then y\(\frac{\pi}{4}\) is equal to
if \(y=\int\frac{e^{\tan x}dx}{\cos^2x(1+e^{2\tan x})}\) and y(0)=6 then y\(\frac{\pi}{4}\) is equal to
Updated On: Apr 10, 2024
- tan-1e-\(\frac{\pi}{4}\)
- tan-1e+6-\(\frac{\pi}{4}\)
- tan-1e-6+\(\frac{\pi}{4}\)
- tan-1\(\frac{1}{e}\)-6+\(\frac{\pi}{4}\)
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
The Correct answer is option is (B) : tan-1e+6-\(\frac{\pi}{4}\)
Was this answer helpful?
0
0
Top Questions on integral
- Find the value of the integral: \[ \int_0^\pi \sin^2(x) \, dx. \]
- Evaluate the integral: $$ \int_0^{\pi/4} \frac{\ln(1 + \tan x)}{\cos x \sin x} \, dx $$
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
- Let $ f(x) $ be a positive function and $I_1 = \int_{-\frac{1}{2}}^1 2x \, f\left(2x(1-2x)\right) dx$ and $I_2 = \int_{-1}^2 f\left(x(1-x)\right) dx.$ Then the value of $\frac{I_2}{I_1}$ is equal to ____
- The value of the definite integral \( \int_0^{\pi} \sin^2 x \, dx \) is:
View More Questions
Questions Asked in JEE Main exam
- 2.8 \( \times 10^{-3} \) mol of \( \text{CO}_2 \) is left after removing \( 10^{21} \) molecules from its ‘\( x \)’ mg sample. The mass of \( \text{CO}_2 \) taken initially is: Given: \( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \)
- JEE Main - 2025
- Mole concept and Molar Masses
- A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is ________________________.
- JEE Main - 2025
- Electromagnetic induction
In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
- JEE Main - 2025
- Electromagnetic Field (EMF)
- The value of \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \) is:
- JEE Main - 2025
- Trigonometric Identities
- Ice at \( -5^\circ C \) is heated to become vapor with temperature of \( 110^\circ C \) at atmospheric pressure. The entropy change associated with this process can be obtained from:
- JEE Main - 2025
- Thermodynamics
View More Questions