Revisions to Simplify an integral involving trigonometric functions

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edited Jan 2, 2017 at 3:27

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We were doing variation of parameters in differential equations tonight and had to do the following integral, the result given by hand calculations.

$$\int \cos t\tan^2 t\,dt=\ln|\sec t+\tan t|-\sin t$$

The students then tried the following in Mathematica:

v2 = Integrate[Cos[t] Tan[t]^2, t]

and got the following result:

-Log[Cos[t/2] - Sin[t/2]] + Log[Cos[t/2] + Sin[t/2]] - Sin[t]

I was able to come home and send them some hand calculated some hand calculated steps to equate this to our solution (minus the absolute value), but I was unable to show them equivalency using Mathematica.

Any suggestions?

We were doing variation of parameters in differential equations tonight and had to do the following integral, the result given by hand calculations.

$$\int \cos t\tan^2 t\,dt=\ln|\sec t+\tan t|-\sin t$$

The students then tried the following in Mathematica:

v2 = Integrate[Cos[t] Tan[t]^2, t]

and got the following result:

-Log[Cos[t/2] - Sin[t/2]] + Log[Cos[t/2] + Sin[t/2]] - Sin[t]

I was able to come home and send them some hand calculated steps to equate this to our solution (minus the absolute value), but I was unable to show them equivalency using Mathematica.

Any suggestions?

We were doing variation of parameters in differential equations tonight and had to do the following integral, the result given by hand calculations.

$$\int \cos t\tan^2 t\,dt=\ln|\sec t+\tan t|-\sin t$$

The students then tried the following in Mathematica:

v2 = Integrate[Cos[t] Tan[t]^2, t]

and got the following result:

-Log[Cos[t/2] - Sin[t/2]] + Log[Cos[t/2] + Sin[t/2]] - Sin[t]

I was able to come home and send them some hand calculated steps to equate this to our solution (minus the absolute value), but I was unable to show them equivalency using Mathematica.

Any suggestions?

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edited Dec 9, 2015 at 3:06

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We were doing variation of parameters in differential equations tonight and had to do the following integral, the result given by hand calculations.

$$\int \cos t\tan^2 t\,dt=\ln|\sec t+\tan t|-\sin t$$

The students then tried the following in MathematicaMathematica:

v2 = Integrate[Cos[t] Tan[t]^2, t]

Andand got the following result.:

-Log[Cos[t/2] - Sin[t/2]] + Log[Cos[t/2] + Sin[t/2]] - Sin[t]

-Log[Cos[t/2] - Sin[t/2]] + Log[Cos[t/2] + Sin[t/2]] - Sin[t]

I was able to come home and send them some hand calculated steps to equate this to our solution (minus the absolute value), but I was unable to show them equivalency using MathematicaMathematica.

Any suggestions?

We were doing variation of parameters in differential equations tonight and had to do the following integral, the result given by hand calculations.

$$\int \cos t\tan^2 t\,dt=\ln|\sec t+\tan t|-\sin t$$

The students then tried the following in Mathematica:

v2 = Integrate[Cos[t] Tan[t]^2, t]

And got the following result.

-Log[Cos[t/2] - Sin[t/2]] + Log[Cos[t/2] + Sin[t/2]] - Sin[t]

I was able to come home and send them some hand calculated steps to equate this to our solution (minus the absolute value), but I was unable to show them equivalency using Mathematica.

Any suggestions?

We were doing variation of parameters in differential equations tonight and had to do the following integral, the result given by hand calculations.

$$\int \cos t\tan^2 t\,dt=\ln|\sec t+\tan t|-\sin t$$

The students then tried the following in Mathematica:

v2 = Integrate[Cos[t] Tan[t]^2, t]

and got the following result:

-Log[Cos[t/2] - Sin[t/2]] + Log[Cos[t/2] + Sin[t/2]] - Sin[t]

I was able to come home and send them some hand calculated steps to equate this to our solution (minus the absolute value), but I was unable to show them equivalency using Mathematica.

Any suggestions?