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mth.122.handout.07

# mth.122.handout.07 - MTH 122 Calculus II Essex County...

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MTH 122 — Calculus II Essex County College — Division of Mathematics and Physics 1 Lecture Notes #7 — Sakai Web Project Material 1 Trigonometric Substitutions, § 7.3 In this section of the textbook we will be presented with difficult integrands that contain a function of x that looks like: p a 2 - x 2 , p a 2 + x 2 , p x 2 - a 2 . In general we will be looking to make a substitution, but in these cases our substitution will be another function with another parameter. To make our calculations simpler, we need to make these new functions invertible, that is, they should be one-to-one . For example, here we’re using a new invertible function g ( θ ) to rewrite the integrand f ( x ): Z f ( x ) d x = Z f ( g ( θ )) g 0 ( θ ) d θ, which really looks like classic u -substitution, and it may even be referred to as inverse u - substitution by some in the mathematical digerati . So let’s take a look at some examples, and I will divide it into three cases. 1. The integrand contains an expression of the form a 2 - x 2 , just use x = a sin θ . Now as you already know, the sine function is not invertible, so we need to restrict the domain of θ [ - π/ 2 , π/ 2], and will use 1 - sin 2 θ = cos 2 θ to help simplify the integral. It should be noted that in general, cos 2 θ = | cos θ | , but for the restricted domain of theta, we have cos 2 θ = cos θ . Example: Z p 1 - 9 x 2 d x = 3 Z p 1 / 9 - x 2 d x Now let x = sin θ 3 , then 3 d x = cos θ d θ .

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mth.122.handout.07 - MTH 122 Calculus II Essex County...

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