Please show me how to get to the answer 1) Evaluate2² sin(x)dxx² – 25dx2) Evaluate

Given, Part a: ∫x2sinxdx Using the By parts rule of the integration, ∫udv=uv-∫vdu Here u=x2 and…

Answered: P sin(a)dz 1) Evaluate Væ² – 25 -da -…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter6: The Trigonometric Functions

Section6.4: Values Of The Trigonometric Functions

Problem 20E

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Question

Please show me how to get to the answer

1) Evaluate
2² sin(x)dx
x² – 25
dx
2) Evaluate

Expert Solution

Step 1

Given,

Part a:

$\int x^{2} \sin (x) d x$

Using the By parts rule of the integration,

$\int u d v = u v - \int v d u$

Here $u = x^{2} and d v = \sin x$

$\Rightarrow - x^{2} \cos (x) - \int - 2 x \cos (x) d x \int - 2 x \cos (x) d x = - 2 (x \sin (x) + \cos (x)) \Rightarrow - x^{2} \cos (x) - [- 2 (x \sin (x) + \cos (x))] \Rightarrow - x^{2} \cos (x) + 2 (x \sin (x) + 2 \cos (x)) \Rightarrow - x^{2} \cos (x) + 2 (x \sin (x) + 2 \cos (x)) + c$

Step 2

Part b:

By applying the trigonometric substitution,

x=5sec(u)

dx=5sec(u)tan(u) du

$\Rightarrow \int \frac{\sqrt{x^{2} - 25}}{x} d x = \int \frac{\sqrt{25 s e c^{2} u - 25}}{s e c u} 5 s e c (u) \tan (u) d u \Rightarrow \int \frac{\sqrt{x^{2} - 25}}{x} d x = \int 5 \sqrt{(s e c^{2} u - 1}) \tan u d u \Rightarrow \int \frac{\sqrt{x^{2} - 25}}{x} d x = \int 5 \sqrt{\tan^{2} u} \tan (u) d u \Rightarrow \int \frac{\sqrt{x^{2} - 25}}{x} d x = \int 5 \tan (u) \tan (u) d u \Rightarrow \int \frac{\sqrt{x^{2} - 25}}{x} d x = \int 5 \tan^{2} u d u$

By using $\int a f (x) d x = a \int f (x) d x$

$\Rightarrow 5 \int \tan^{2} (u) d u$

$\tan^{2} (u) + 1 = s e c^{2} (u) \Rightarrow 5 \int - 1 + s e c^{2} (u) d u$

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P sin(a)dz 1) Evaluate Væ² – 25 -da - ,. 2) Evaluate