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Chapter 7: Problem 3

What trigonometric identity is useful in evaluating \(\int \sin ^{2} x d x ?\)

Short Answer

Expert verified
Answer: The useful trigonometric identity is \(\sin^2(x) = \frac{1 - \cos(2x)}{2}\). The final integral expression is \(\int \sin^2(x) dx = \frac{1}{2}x - \frac{1}{4}\sin(2x) + C\).

Step by step solution

01

Recall the double angle identity for cosine function

The double angle identity for the cosine function is as follows: $$\cos(2x) = 1 - 2\sin^2(x)$$
02

Solve for \(\sin^2(x)\)

Now, we'll solve the above identity for \(\sin^2(x)\). This will give us the trigonometric identity we need in order to simplify the integrand: $$\sin^2(x) = \frac{1 - \cos(2x)}{2}$$
03

Replace \(\sin^2(x)\) in the integral

Now that we have the trigonometric identity for \(\sin^2(x)\), plug it into the integral and simplify: $$\int \sin^2(x) dx = \int \frac{1 - \cos(2x)}{2} dx$$
04

Split the integral

Now, split the integral into two separate integrals: $$\int \frac{1 - \cos(2x)}{2} dx = \frac{1}{2}\int1 dx - \frac{1}{2}\int\cos(2x) dx$$
05

Integrate

Now, integrate the two integrals separately: $$\frac{1}{2}\int1 dx - \frac{1}{2}\int\cos(2x) dx = \frac{1}{2}x - \frac{1}{4}\sin(2x) + C$$ Here, \(C\) is the constant of integration. So, the trigonometric identity that is useful in evaluating the given integral is: $$\sin^2(x) = \frac{1 - \cos(2x)}{2}$$ And the integral of \(\sin^2(x)\) with respect to \(x\) is: $$\int \sin^2(x) dx = \frac{1}{2}x - \frac{1}{4}\sin(2x) + C$$

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Most popular questions from this chapter

Evaluate the following integrals. Assume a and b are real numbers and \(n\) is an integer. $$\int \frac{x}{a x+b} d x \text { (Use } u=a x+b$$

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\(A n\) integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or equivalently \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. \(A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}\) $$\text { Evaluate } \int_{0}^{\pi / 2} \frac{d \theta}{\cos \theta+\sin \theta}$$

Show that \(L=\lim _{n \rightarrow \infty}\left(\frac{1}{n} \ln n !-\ln n\right)=-1\) in the following steps. a. Note that \(n !=n(n-1)(n-2) \cdots 1\) and use \(\ln (a b)=\ln a+\ln b\) to show that $$ \begin{aligned} L &=\lim _{n \rightarrow \infty}\left(\left(\frac{1}{n} \sum_{k=1}^{n} \ln k\right)-\ln n\right) \\ &=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \frac{k}{n} \end{aligned} $$ b. Identify the limit of this sum as a Riemann sum for \(\int_{0}^{1} \ln x d x\) Integrate this improper integral by parts and reach the desired conclusion.

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