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Sindrei [870]
3 years ago
14

A. Evaluate ∫20 tan 2x sec^2 2x dx using the substitution u = tan 2x.

Mathematics
1 answer:
irakobra [83]3 years ago
3 0

Answer:

The integral is equal to 5\sec^2(2x)+C for an arbitrary constant C.

Step-by-step explanation:

a) If u=\tan(2x) then du=2\sec^2(2x)dx so the integral becomes \int 20\tan(2x)\sec^2(2x)dx=\int 10\tan(2x) (2\sec^2(2x))dx=\int 10udu=\frac{u^2}{2}+C=10(\int udu)=10(\frac{u^2}{2}+C)=5\tan^2(2x)+C. (the constant of integration is actually 5C, but this doesn't affect the result when taking derivatives, so we still denote it by C)

b) In this case u=\sec(2x) hence du=2\tan(2x)\sec(2x)dx. We rewrite the integral as \int 20\tan(2x)\sec^2(2x)dx=\int 10\sec(2x) (2\tan(2x)\sec(2x))dx=\int 10udu=5\frac{u^2}{2}+C=5\sec^2(2x)+C.

c) We use the trigonometric identity \tan(2x)^2+1=\sec(2x)^2 is part b). The value of the integral is 5\sec^2(2x)+C=5(\tan^2(2x)+1)+C=5\tan^2(2x)+5+C=5\tan^2(2x)+C. which coincides with part a)

Note that we just replaced 5+C by C. This is because we are asked for an indefinite integral. Each value of C defines a unique antiderivative, but we are not interested in specific values of C as this integral is the family of all antiderivatives. Part a) and b) don't coincide for specific values of C (they would if we were working with a definite integral), but they do represent the same family of functions.  

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