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Chapter 7: Problem 21

Evaluate the following integrals. $$\int x \sin x \cos x d x$$

Short Answer

Expert verified
Question: Evaluate the integral \(\int x \sin x \cos x d x\). Answer: \(-\frac{1}{4}x\cos{2x} +\frac{1}{8}\sin{2x} + C\)

Step by step solution

01

Identify Integration by Parts components

Since we're unable to integrate the given function directly, we must identify which components we need to integrate by parts. Integration by parts formula states: $$\int u dv = uv - \int v du.$$ In this problem, let $$u=x,\qquad dv = \sin x \cos x dx.$$
02

Compute du and v

We will now differentiate u with respect to x and find the antiderivative of dv. From \(u=x\), we have $$du = dx.$$ Now, we can use the double-angle identity of sine to simplify the integration of dv. $$\sin 2a = 2\sin a \cos a$$ $$2 \sin{\frac{x}{2}}\cos{\frac{x}{2}}=\sin x\cos x$$ We can now write our equation for dv as: $$dv = \frac{1}{2}\sin 2x dx$$ Integrating dv with respect to x: $$v=\int dv $$ $$v= \int \frac{1}{2}\sin 2x dx$$ $$v= -\frac{1}{4}\cos 2x + C.$$
03

Apply Integration by Parts

Now that we have computed \(u\), \(du\), \(dv\), and \(v\), we can use the integration by parts formula. $$ \int x \sin x \cos x d x = \int u dv $$ According to the integration by parts formula: $$\int u dv = uv - \int v du$$ Substitute the u, du, dv, and v values into the formula: $$\int x \sin x \cos x d x = -\frac{1}{4}x\cos{2x} +\frac{1}{4}\int \cos{2x} dx$$
04

Integrate the remaining integral

Now integrate the remaining integral: $$\int \cos{2x} dx = \frac{1}{2}\sin{2x} + C$$
05

Combine all terms

Combine all the terms to get the final answer: $$\int x \sin x \cos x d x = -\frac{1}{4}x\cos{2x} +\frac{1}{8}\sin{2x} + C$$

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