How do you evaluate the integral #int tan^3theta#?

Answer 1

#int tan^3x dx = 1/2tan^2x +ln abs(cosx) + C#

Use the trigonometric identity:

#tan^2x = sec^2x-1#

to get:

#int tan^3x dx = int (sec^2x -1) tanx dx= int tanxsec^2xdx -int tanx dx#
Solve the first integral using: #d(tanx) = sec^2xdx#
#int tanx sec^2 dx = int tanx d(tanx) = 1/2tan^2x +C_1#

For the second integral:

#int tanx dx = int sinx/cosx dx = - int (d(cosx))/cosx = -ln abs(cosx) + C_2#

Putting it together:

#int tan^3x dx = 1/2tan^2x +ln abs(cosx) + C#
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Answer 2

To evaluate the integral ∫tan^3(θ)dθ, we can use trigonometric identities and integration by parts. By rewriting tan^3(θ) as (sec^2(θ) - 1)tan(θ), we can separate it into two integrals. Then, using the substitution u = tan(θ), du = sec^2(θ)dθ, we can solve each integral separately. Integrating sec^2(θ) yields tan(θ), and integrating 1 yields θ. Therefore, the integral of tan^3(θ) is equal to (1/2)tan^2(θ) - tan(θ) + θ + C, where C is the constant of integration.

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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.

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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