/*! 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 10 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 10

Evaluate the following integrals. $$\int \sin ^{3} x d x$$

Short Answer

Expert verified
Question: Evaluate the integral of the function sin^3(x) with respect to x. Answer: The evaluated integral of sin^3(x) is given by: \(\frac{-3\cos x + \cos^3 x + 3C}{3} + C_1\).

Step by step solution

01

Rewrite the integrand using identities

: We can rewrite \(\sin^3(x)\) as \(\sin(x) \cdot \sin^2(x)\). From the properties of trigonometric functions, we know that \(\sin^2(x) = 1 - \cos^2(x)\). So now, our integral becomes: $$\int \sin(x) (1 - \cos^2(x)) dx$$
02

Expand the integrand and integrate terms

: Now let's expand the integrand and integrate each term separately: $$\int \sin(x) (1 - \cos^2(x)) dx = \int \sin x dx - \int \sin x \cos^2(x) dx$$ We know that the integral of \(\sin x\) is \(-\cos x + C\). For the second term, we can use the substitution method. Let \(u = \cos x\), then \(du = -\sin x dx\). So, the integral becomes: $$-\int \cos^2u (-du)$$ Now we have: $$-\int (1 - u^2) du = -u + \frac{u^3}{3} + C_1$$ Substitute back with \(\cos x\) for \(u\), we get: $$-\cos x + \frac{\cos^3 x}{3} + C_1$$ Putting the integrals together, the final answer is: $$\left(-\cos x + C\right) - \left(-\cos x + \frac{\cos^3 x}{3} + C_1\right) = \frac{-3\cos x + \cos^3 x + 3C}{3} + C_1$$

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