How do you find the indefinite integral of #int (1+secpix)^2secpixtanpixdx#?
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To find the indefinite integral of ( \int (1+\sec(\pi x))^2 \sec(\pi x) \tan(\pi x) , dx ), you can use the substitution method. Let ( u = 1 + \sec(\pi x) ), then ( du = \pi \sec(\pi x) \tan(\pi x) , dx ). Rearranging, we get ( \frac{du}{\pi} = \sec(\pi x) \tan(\pi x) , dx ).
Substitute ( u ) and ( \frac{du}{\pi} ) into the integral: [ \int u^2 , \frac{du}{\pi} ]
Now, integrate with respect to ( u ): [ \frac{1}{\pi} \int u^2 , du = \frac{1}{\pi} \left( \frac{u^3}{3} + C \right) = \frac{1}{3\pi} (1 + \sec(\pi x))^3 + C ]
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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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