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Chapter 5: Q. 57 (page 496)

${鈭�}_{0}^{\frac{蟺}{2}} \tan^{3} x s e c x d x$

Short Answer

Expert verified

Undefined.

The limit does not converge.

Step by step solution

01

Given

${鈭�}_{0}^{\frac{蟺}{2}} \tan^{3} x s e c x d x$

02

Calculate integral

As we know,

$1 + \tan^{2} x = s e c^{2} x 鈬� \tan^{2} x = s e c^{2} x - 1$

So that,

${鈭�}_{0}^{\frac{蟺}{2}} \tan^{3} x s e c x d x = {鈭�}_{0}^{\frac{蟺}{2}} \tan x s e c x (s e c^{2} x - 1) d x$

Let,

$u = s e c x d u = s e c x \tan x d x d x = \frac{d u}{s e c x \tan x}$

Define the limit of integration,

$x = 0 鈬� u = s e c 0 = 1 x = \frac{蟺}{2} 鈬� u = s e c \frac{蟺}{2} = 鈭�$

Therefore,

${鈭�}_{0}^{\frac{蟺}{2}} \tan x s e c x (s e c^{2} x - 1) d x = {鈭�}_{1}^{鈭�} (u^{2} - 1) d u = {[\frac{u^{3}}{3} - u]}_{1}^{鈭�} = 鈭� (Undefined)$

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Most popular questions from this chapter

Complete the square for each quadratic in Exercises 28鈥�33. Then describe the trigonometric substitution that would be appropriate if you were solving an integral that involved that quadratic.

$2 x^{2} 鈭� 4 x + 1$

Find three integrals in Exercises 27鈥�70 for which a good strategy is to apply integration by parts twice.

Find three integrals in Exercises 27鈥�70 for which a good strategy is to use integration by parts with $u = x$ and dv the remaining part.

Show by differentiating (and then using algebra) that $鈭� \cot 鈦� (\sin^{鈭� 1} 鈦� x)$ and $鈭� \frac{\sqrt{1 鈭� x^{2}}}{x}$ are both antiderivatives of $\frac{1}{x^{2} \sqrt{1 鈭� x^{2}}}$ . How can these two very different-looking functions be an antiderivative of the same function?

Write $\sin (2 \cos^{- 1} x)$ as an algebraic function.

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