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Evaluate the indefinite integral: #int sin 35t sec^2 (cos 35t)dt#?

Answer 1

Ezekiel Alexander

#-1/35tan(cos35t)+C#

#intsin35tsec^2(cos35t)dt#

Use the substitution #u=cos35t=>du=-35sin35t#.

#=-1/35intsec^2(cos35t)(-35sin35t)dt#

#=-1/35intsec^2(u)du#

This is a common integral:

#=-1/35tan(u)+C#

#=-1/35tan(cos35t)+C#

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Answer 2

Emily Ammerman

To evaluate the indefinite integral ( \int \sin(35t) \sec^2(\cos(35t)) , dt ), you can use a substitution method. Let ( u = \cos(35t) ), then ( du = -35 \sin(35t) , dt ). This gives:

[ \int \sin(35t) \sec^2(\cos(35t)) , dt = \int -\frac{1}{35} \sec^2(u) , du ]

Using the integral of ( \sec^2(u) ), which is ( \tan(u) + C ), where ( C ) is the constant of integration, we get:

[ \int -\frac{1}{35} \sec^2(u) , du = -\frac{1}{35} \tan(u) + C ]

Substituting back ( u = \cos(35t) ), we get:

[ = -\frac{1}{35} \tan(\cos(35t)) + C ]

This is the antiderivative of the given integral.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.