Question:

Evaluate: $\int sec^{4/3} \,x cosec^{8/3}\,x \,dx$

Updated On: Mar 6, 2025

$ \frac{3}{5} tan^{-5/3} x - 3 tan^{1/3} x + C $
$ - \frac{3}{5} tan^{-5/3} x - 3 tan^{1/3} x + C $
$- \frac{3}{5} tan^{-5/3} x - 3 tan^{1/3} x + C $
None of these

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The Correct Option is B

Solution and Explanation

Let $ I= \int sec^{4/3} x \,cosec^{8/3} x \,dx$ $\Rightarrow I= \int cos^{-4/3} x \,sin^{-8/3} x \,dx$ $\Rightarrow I= \int \frac{sec^{4} \,x}{tan^{\frac{8}{3}} \,x}dx = \int \frac{\left(1+tan^{2} x\right)}{tan^{\frac{8}{3}} x} sec^{2} \,x\, dx$ Putting $tanx = t \Rightarrow sec^{2}\,x\, dx = dt$ $I = \int \frac{1+t^{2}}{t^{\frac{8}{3}}} dt = \int \left(t^{-\frac{8}{3}} + t^{-\frac{2}{3}}\right) dt$ $\Rightarrow I=\frac{-3}{5}t^{-\frac{5}{3}} + 3t^{\frac{1}{3}} + C$ $= \frac{-3}{5} tan^{-5/3} x + 3 tan^{1/3} x + C $

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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.