The solution of
$\frac{dy}{dx} = \frac{y}{x}+\tan \frac{y}{x}$ is
- x = c sin (y/x)
- x = c sin (xy)
- y = c sin (y/x)
- xy = c sin (x/y)
The Correct Option is A
Solution and Explanation
Top Questions on integral
- The value of the integral \( \int_0^1 x^2 \, dx \) is:
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 ____
- Evaluate the integral: \[ \int \frac{x^2 + 2x}{\sqrt{x^2 + 1}} \, dx \]
- Evaluate the integral: \[ \int \sqrt{x^2 + 3x} \, dx \]
Questions Asked in WBJEE exam
What are the charges stored in the \( 1\,\mu\text{F} \) and \( 2\,\mu\text{F} \) capacitors in the circuit once current becomes steady?
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Manufacturers supply a zener diode with zener voltage \( V_z=5.6\,\text{V} \) and maximum power dissipation \( P_{\max}=\frac14\,\text{W} \). This zener diode is used in the circuit shown. Calculate the minimum value of the resistance \( R_s \) so that the zener diode will not burn when the input voltage is \( V_{in}=10\,\text{V} \).
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Two charges \( +q \) and \( -q \) are placed at points \( A \) and \( B \) respectively which are at a distance \( 2L \) apart. \( C \) is the midpoint of \( AB \). The work done in moving a charge \( +Q \) along the semicircle CSD (\( W_1 \)) and along the line CBD (\( W_2 \)) are
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A piece of granite floats at the interface of mercury and water. If the densities of granite, water and mercury are \( \rho, \rho_1, \rho_2 \) respectively, the ratio of volume of granite in water to that in mercury is
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.