How do you integrate $\int (θ^{2} + {sec}^{2} θ) d θ$ $int (theta^2+sec^2theta)d theta$ ?

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How do you integrate $\int (θ^{2} + {sec}^{2} θ) d θ$ $int (theta^2+sec^2theta)d theta$ ?

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Answer 1 · 2016-12-05T21:26:26

$\frac{θ^{3}}{3} + tan θ + C$ $theta^3/3+tantheta+C$

Explanation:

$I = \int (θ^{2} + {sec}^{2} θ) d θ$ $I=int(theta^2+sec^2theta)d theta$

Split up the integral:

$I = \int θ^{2} d θ + \int {sec}^{2} θ d θ$ $I=inttheta^2d theta+intsec^2thetad theta$

Think of the following two things:

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx=x^(n+1)/(n+1)+C$
$\frac{d}{d x} tan x = {sec}^{2} x \Rightarrow \int {sec}^{2} x d x = tan x + C$ $d/dxtanx=sec^2x=>intsec^2xdx=tanx+C$

Don't be tripped up by the use of $θ$ $theta$ as a variable—this process is identical to integrating $\int (x^{2} + {sec}^{2} x) d x$ $int(x^2+sec^2x)dx$ .

Thus:

$I = \frac{θ^{3}}{3} + tan θ + C$ $I=theta^3/3+tantheta+C$

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How do you integrate $\int (θ^{2} + {sec}^{2} θ) d θ$ $int (theta^2+sec^2theta)d theta$ ?

1 Answer

Explanation:

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How do you integrate int (theta^2+sec^2theta)d theta∫(θ2+sec2θ)dθint (theta^2+sec^2theta)d theta?

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How do you integrate $\int (θ^{2} + {sec}^{2} θ) d θ$ $int (theta^2+sec^2theta)d theta$ ?