f(𝜃) = 3 sin(𝜃) − 4 sec(𝜃) tan(𝜃)&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;on the interval (-𝜋/2 ,𝜋/2)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

Question

f(𝜃) = 3 sin(𝜃) − 4 sec(𝜃) tan(𝜃)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; &amp;nbsp;on the interval (-𝜋/2 ,𝜋/2)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;

nick1337 · Accepted Answer

3sin(θ)-4sec(θ)tan(θ)Write&amp;nbsp;3sin(θ)-4sec(θ)tan(θ)&amp;nbsp;as a function.f(θ)=3sin(θ)-4sec(θ)tan(θ)The function&amp;nbsp;F(θ)&amp;nbsp;can be found by finding the indefinite integral of the derivative&amp;nbsp;f(θ).F(θ)=∫f(θ)dθSet up the integral to solve.F(θ)=∫3sin(θ)-4sec(θ)tan(θ)dθSplit the single integral into multiple integrals.∫3sin(θ)dθ+∫-4sec(θ)tan(θ)dθSince&amp;nbsp;3&amp;nbsp;is constant with respect to&amp;nbsp;θ, move&amp;nbsp;3&amp;nbsp;out of the integral.3∫sin(θ)dθ+∫-4sec(θ)tan(θ)dθThe integral of&amp;nbsp;sin(θ)&amp;nbsp;with respect to&amp;nbsp;θ&amp;nbsp;is&amp;nbsp;-cos(θ).3(-cos(θ)+C)+∫-4sec(θ)tan(θ)dθSince&amp;nbsp;-4&amp;nbsp;is constant with respect to&amp;nbsp;θ, move&amp;nbsp;-4&amp;nbsp;out of the integral.3(-cos(θ)+C)-4∫sec(θ)tan(θ)dθSince the derivative of&amp;nbsp;sec(θ)&amp;nbsp;is&amp;nbsp;sec(θ)tan(θ), the integral of&amp;nbsp;sec(θ)tan(θ)&amp;nbsp;is&amp;nbsp;sec(θ).3(-cos(θ)+C)-4(sec(θ)+C)Simplify.-3cos(θ)-4sec(θ)+CThe answer is the antiderivative of the function&amp;nbsp;f(θ)=3sin(θ)-4sec(θ)tan(θ).F(θ)=−3cos(θ)−4sec(θ)+C

f(𝜃) = 3 sin(𝜃) − 4 sec(𝜃)

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Answer & Explanation

New Questions in Calculus 1