The value of the integral \(∫^{ \frac{π}{2}}_{ \frac{-π}{2}} \frac{dx}{(1+e^x)(sin^6x+cos^6x)}\) is equal to
- \(2π\)
- 0
- \(π\)
- \(\frac{π}{2}\)
The Correct Option is C
Solution and Explanation
\(I\) = \(∫^{ \frac{π}{2}}_{ \frac{-π}{2}} \frac{dx}{(1+e^x)(sin^6x+cos^6x)}\).....(i)
\(I\) = \(∫^{ \frac{π}{2}}_{ \frac{-π}{2}} \frac{dx}{(1+e^{-x})(sin^6x+cos^6x)}\).....(ii)
From equation (i) & (ii)
\(2I\) = \(∫ ^{\frac{π}{2}} _{\frac{-π}{2}} \frac{dx}{(sin^6x+cos^6x) }\) = \(∫ ^{\frac{π}{2}}_0 \frac{ dx}{(1-\frac{3}{4}sin^22x)}\)
\(⇒ I = ∫ ^{\frac{π}{2}}_ 0 \frac{4sec^22xdx}{(4+tan^22x)} = 2 ∫^{\frac{π}{4}} _0\frac{ 4sec^22x}{4+tan^22x }dx\)
Now,
\(tan2x = t\) and \(2sec^22xdx = dt\)
At \(x = 0, t = 0\)
is \(x = \frac{π}{4}, t →∞\)
\(∴ I = 2 ∫^∞_0 \frac{2dt}{4+t^2} = 2(tan^{-1} \frac{t}{2})^∞_0\)
= \(2 \frac{π}{2} = π\)
Hence, the correct option is (C): \(\pi\)
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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.