/*! 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 25 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 25

Evaluate the following integrals. $$\int \frac{\sin t+\tan t}{\cos ^{2} t} d t$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int \frac{\sin t+\tan t}{\cos ^{2} t} d t$$. Answer: $$\frac{1}{\cos t} + \frac{1}{2\cos^2 t} - t + C$$.

Step by step solution

01

Separate the integral into simpler parts

We will first separate the integral into two parts in order to make it easier to solve: $$\int \frac{\sin t+\tan t}{\cos ^{2} t} d t= \int \frac{\sin t}{\cos ^{2} t}d t + \int \frac{\tan t}{\cos ^{2} t}d t$$
02

Rewrite the integrals using trigonometric identities

We can rewrite the integrals using some trigonometric identities that will make them easier to solve: $$\int \frac{\sin t}{\cos ^{2} t}d t = \int \frac{1 - \cos^2 t}{\cos^2 t} dt = \int \frac{1}{\cos^2 t} dt - \int dt$$ And $$\int \frac{\tan t}{\cos^{2}t} d t = \int \frac{\sin t}{\cos t \cdot \cos t} d t = \int \frac{\sin t}{\cos^3 t} d t$$
03

Solve the first integral

Now let's solve the first integral using the substitution method. Make the substitution \(u = \cos t\) and \(du=-\sin t dt\). This results in: $$\int \frac{1}{\cos^2 t} dt - \int dt =\int \frac{1}{u^2}(-du) - \int dt =-\int \frac{1}{u^2} du - \int dt$$ Now, integrate both terms: $$-\int \frac{1}{u^2} du - \int dt = -\int u^{-2} du - \int dt$$ This gives us: $$-[-1 \cdot u^{-1}] - t + C = \frac{1}{u} - t + C = \frac{1}{\cos t} - t + C_1$$ where \(C_1\) is a constant.
04

Solve the second integral

To solve the second integral, we can use the substitution method again. Make the substitution \(v = \cos t\) and \(dv = -\sin t dt\). This results in: $$\int \frac{\sin t}{\cos^3 t} d t = -\int \frac{1}{v^3} dv$$ Now, integrate this term: $$-\int \frac{1}{v^3} dv = -\int v^{-3} dv = -[-\frac{1}{2}v^{-2}] + C_2 = \frac{1}{2v^2} + C_2 = \frac{1}{2\cos^2 t} + C_2$$ where \(C_2\) is a constant.
05

Add the results of the two integrals

Now, we can add the results of the two integrals together: $$\int \frac{\sin t+\tan t}{\cos ^{2} t} d t = \left(\frac{1}{\cos t} - t + C_1\right) + \left(\frac{1}{2\cos^2 t} + C_2\right) = \frac{1}{\cos t} + \frac{1}{2\cos^2 t} - t + C$$ where \(C = C_1 + C_2\) is a constant.

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