If \(∫\frac{1}{x}\) \(√{\frac{1-x}{1+x}}\) dx = \(g(x) + c,g(1) = 0\) , then g \((\frac{1}{2})\)
is equal to
If \(∫\frac{1}{x}\) \(√{\frac{1-x}{1+x}}\) dx = \(g(x) + c,g(1) = 0\) , then g \((\frac{1}{2})\)
is equal to
\(log_e(\frac{√3-1}{√3+1})\)\(+\frac{π}{3}\)
\(log_e(\frac{√3+1}{√3-1})\)\(+\frac{π}{3}\)
\(log_e(\frac{√3+1}{√3-1})\)\(-\frac{π}{3}\)
\(\frac{1}{2}log_e(\frac{√3-1}{√3+1})-\)\(\frac{π}{6}\)
The Correct Option is A
Solution and Explanation
The correct answer is (A) : \(log_e(\frac{√3-1}{√3+1})\)\(+\frac{π}{3}\)
∵ \(∫\frac{1}{x}\) \(√{\frac{1-x}{1+x}}\) \(dx = g(x) + c\)
\(\int_1^{\frac{1}{2}}\frac{1}{x} \sqrt{\frac{1-x}{1+x}} \,dx = g\left(\frac{1}{2}\right) - g(1)\)
∴\( g(\frac{1}{2}) =\) \(\int_1^{\frac{1}{2}}\frac{1}{x} \sqrt{\frac{1-x}{1+x}} \,dx\)
\(cotx=cos2θ\)
= \(\int_0^{\frac{π}{6}} \frac{1}{cos2θ}.\frac{sinθ}{cosθ}(sinθ/cosθ ( -2sin2θ)dθ\)
= \(-\int_0^{\frac{π}{6}}\frac{4sin²θ}{cos2θ}dθ\)
= \(2\int_0^{\frac{π}{6}}\)\(\frac{(1 - 2sin²θ -1)}{cos2θ }\)\(dθ\)
= \(2 \int^{\frac{π}{6}}_0 \)\(( 1 - sec2θ ) dθ\)
=\( \frac{π}{3}\) \(- 2. \frac{1}{2} \)\([ In | sec2θ + tan2θ| ]^{\frac{π}{6}}_0\)
= \( \frac{π}{3}\) \(- [ In | 2 + √3 | - In1 ] \)
= \( \frac{π}{3}\) \(+ In ( \frac{1}{2 }+ √3 )\)
= \( \frac{π}{3}\) \(+ In | \frac{√3 - 1}{√3 + 1} |\)
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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.