/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 27

Evaluate the following integrals. $$\int \frac{2-3 x}{\sqrt{1-x^{2}}} d x$$

Short Answer

Expert verified
The antiderivative of the integrand is \(2\arcsin{x} + 3\cos{(\arcsin{x})} + C\).

Step by step solution

01

Make the Substitution

Let \(x = \sin{u}\). Then, \(dx = \cos{u} du\). Now, substitute this expression for \(x\) and \(dx\) in the integral: $$\int \frac{2-3\sin{u}}{\sqrt{1-\sin^2{u}}} \cos{u} du$$
02

Simplify the Integral

Now we can simplify the integrand using the identity \(1 - \sin^2{u} = \cos^2{u}\): $$\int \frac{2-3\sin{u}}{\sqrt{\cos^2{u}}} \cos{u} du = \int \frac{2-3\sin{u}}{\cos{u}} \cos{u} du$$ The \(\cos{u}\) terms cancel out, leaving: $$\int (2-3\sin{u}) du$$
03

Integrate

Now we can integrate the remaining expression with respect to \(u\): $$\int (2-3\sin{u}) du = 2\int du - 3\int \sin{u} du = 2u - 3(-\cos{u}) + C = 2u + 3\cos{u} + C$$
04

Substitute Back

Now that we have integrated the expression, we need to substitute back the original variable \(x\). Recall that we had \(x = \sin{u}\), so \(u = \arcsin{x}\). Therefore: $$2u + 3\cos{u} + C = 2\arcsin{x} + 3\cos{(\arcsin{x})} + C$$ We have now found the antiderivative of the given integrand.
05

Final Answer:

\(2\arcsin{x} + 3\cos{(\arcsin{x})} + C\)

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Most popular questions from this chapter

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