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Chapter 4: Problem 8

Evaluate \(\int \cos a x d x\) and \(\int \sin a x d x,\) where \(a\) is a constant.

Short Answer

Expert verified
The integrals are: 1. For cos(ax)dx: \( \int\cos(ax)dx = \frac{1}{a}\sin(ax) + C_1\) 2. For sin(ax)dx: \(\int\sin(ax)dx = -\frac{1}{a}\cos(ax) + C_2\)

Step by step solution

01

Evaluate \(\int \cos ax dx\)

To evaluate this integral, we'll use the substitution technique. Let us substitute \(u=ax\). Therefore, \(du = a dx\) or \(dx = \frac{1}{a}du.\) Now, the integral becomes $$\int \cos ax dx = \int \cos(u)\frac{1}{a} du$$ This can now be integrated directly: $$\frac{1}{a}\int \cos(u) du$$
02

Integrate \(\cos(u)\) with respect to \(u\)

Recall that the integral of \(\cos(u)\) with respect to \(u\) is \(\sin(u)\). Therefore, we have: $$\frac{1}{a}\int \cos(u) du = \frac{1}{a}\sin(u) + C_1$$ where \(C_1\) is an integration constant.
03

Replace \(u\) with \(ax\)

Now, we replace \(u\) with the original variable by substituting \(ax\) back in: $$\frac{1}{a}\sin(u) + C_1 = \frac{1}{a}\sin(ax) + C_1$$ So, the evaluated integral of \(\cos ax dx\) is: $$\int\cos(ax)dx = \frac{1}{a}\sin(ax) + C_1$$
04

Evaluate \(\int \sin ax dx\)

We will use the same substitution technique for this integral. Again, let us substitute \(v = ax\). Therefore, \(dv = adx\) or \(dx = \frac{1}{a}dv.\) Now, the integral becomes $$\int \sin ax dx = \int \sin(v)\frac{1}{a} dv$$ This can now be integrated directly: $$\frac{1}{a}\int \sin(v) dv$$
05

Integrate \(\sin(v)\) with respect to \(v\)

Recall that the integral of \(\sin(v)\) with respect to \(v\) is \(-\cos(v)\). Therefore, we have: $$\frac{1}{a}\int \sin(v) dv = -\frac{1}{a}\cos(v) + C_2$$ where \(C_2\) is another integration constant.
06

Replace \(v\) with \(ax\)

Now, we replace \(v\) with the original variable by substituting \(ax\) back in: $$-\frac{1}{a}\cos(v) + C_2 = -\frac{1}{a}\cos(ax) + C_2$$ So, the evaluated integral of \(\sin ax dx\) is: $$\int\sin(ax)dx = -\frac{1}{a}\cos(ax) + C_2$$ Finally, we have the two evaluated integrals: 1. \(\int\cos(ax)dx = \frac{1}{a}\sin(ax) + C_1\) 2. \(\int\sin(ax)dx = -\frac{1}{a}\cos(ax) + C_2\)

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