How do you find the integral of #tan^3(2x) sec^100(2x) dx#?
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To find the integral of ( \tan^3(2x) \sec^{100}(2x) , dx ), you can use the substitution method. Let ( u = \tan(2x) ). Then, ( du = 2\sec^2(2x) , dx ).
Rewrite the integral in terms of ( u ):
[ \int \tan^3(2x) \sec^{100}(2x) , dx = \frac{1}{2} \int u^3 \sec^{98}(2x) , du ]
Now, we can use another substitution. Let ( t = \sec(2x) ). Then, ( dt = 2\tan(2x) \sec(2x) , dx ).
We can express ( \tan(2x) ) in terms of ( t ):
[ \tan(2x) = \frac{u}{\sqrt{1 + u^2}} ]
Now, substitute ( u = t ) and ( du = 2\tan(2x) \sec(2x) , dx = 2t , dt ):
[ \frac{1}{2} \int \frac{t^3}{\sqrt{1 + t^2}} \sec^{98}(2x) , (2t , dt) = \int \frac{t^4}{\sqrt{1 + t^2}} \sec^{98}(2x) , dt ]
This integral can be solved using integration by parts, or you may explore other methods depending on the specific requirements of your problem.
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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.
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.
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.
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.

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