# How do you evaluate the integral #(1+tan(x))^3 * sec^2(x)dx# within the range [0,pi/4]?

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To evaluate the integral ( \int_0^{\frac{\pi}{4}} (1+\tan(x))^3 \cdot \sec^2(x) , dx ) within the range ([0, \frac{\pi}{4}]), you can use the substitution method.

- Let ( u = 1 + \tan(x) ). Then ( du = \sec^2(x) , dx ).
- When ( x = 0 ), ( u = 1 + \tan(0) = 1 ).
- When ( x = \frac{\pi}{4} ), ( u = 1 + \tan(\frac{\pi}{4}) = 1 + 1 = 2 ).
- Rewrite the integral in terms of ( u ): ( \int_1^2 u^3 , du ).
- Integrate ( u^3 ) with respect to ( u ) to get ( \frac{u^4}{4} ).
- Evaluate the integral from ( 1 ) to ( 2 ).
- Substitute ( 2^4/4 - 1^4/4 ) for ( u^4/4 ).
- Calculate ( \frac{2^4}{4} - \frac{1^4}{4} = 4 - \frac{1}{4} = \frac{15}{4} ).

Therefore, the value of the integral ( \int_0^{\frac{\pi}{4}} (1+\tan(x))^3 \cdot \sec^2(x) , dx ) within the range ([0, \frac{\pi}{4}]) is ( \frac{15}{4} ).

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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.

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