Question

quiz 12 - requires respondus lockdown browser + webcam
started nov 9 at 5:13pm
quiz instructions
access code: start
timed: 25 minutes
number of attempts: 1
lockdown browser and respondus monitor required
this quiz covers material of this module through 5.4.
question 3
find the general indefinite integral \\( \int\left(2+\tan ^{2} \theta\
ight) d \theta \\)
\\( \theta+\tan \theta+c \\)
\\( 4 \theta-\sec \theta+c \\)
\\( 3 \theta-\tan \theta+c \\)
\\( 2 \theta+\sec \theta+c \\)

Explanation:

Step1: Use trigonometric identity

Recall the identity \(1+\tan^{2}\theta=\sec^{2}\theta\), so \(2 + \tan^{2}\theta=1+(1 + \tan^{2}\theta)=1+\sec^{2}\theta\).
Then \(\int(2+\tan^{2}\theta)d\theta=\int(1 + \sec^{2}\theta)d\theta\).

Step2: Integrate term - by - term

We know that \(\int 1d\theta=\theta+C_1\) and \(\int\sec^{2}\theta d\theta=\tan\theta + C_2\) (where \(C_1,C_2\) are constants of integration).
By the sum rule of integration \(\int(f(x)+g(x))dx=\int f(x)dx+\int g(x)dx\), we have \(\int(1+\sec^{2}\theta)d\theta=\int 1d\theta+\int\sec^{2}\theta d\theta\).

Answer:

\(\theta+\tan\theta + C\) (where \(C = C_1 + C_2\) is the constant of integration)