Find the following integral: \(\int \frac{\sec^2 x}{\cosec^2 x}dx\)
Solution and Explanation
\(\int \frac{\sec^2 x}{\cosec^2 x}dx\)
= \(\int \frac{\frac{1}{\cos^2 x}}{\frac{1}{\sin^2 x}}dx\)
= \(\int \frac{\sin^2 x}{\cos^2 x}dx\)
=\(\int \tan^2 x dx\)
= \(\int (\sec^2 x-1)dx\)
= \(\int \sec^2 xdx - \int 1dx\)
= \(\tan x -x+C\)
Top Questions on integral
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Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
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- If \( \int \frac{1}{2x^2} \, dx = k \cdot 2x + C \), then \( k \) is equal to:
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- electrostatic potential and capacitance
- The diagonals of a parallelogram are given by \( \mathbf{a} = 2 \hat{i} - \hat{j} + \hat{k} \) and \( \mathbf{b} = \hat{i} + 3 \hat{j} - \hat{k}\) . Find the area of the parallelogram.
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.