Integrate tan^3x dx (please check my soln method)

[marian51]

I know that the method I used to try to solve this integral isn't the best method, but I'm curious to know if it is even an acceptable way to try to solve the integral.

Since the integral was tan^3(x) I began by drawing a right triange with an angle of x. The side adjacent to angle x I termed 1, the side opposite angle x I termed y. Therefore based on this triangle I set tanx=y and therefore tan^3x=y^3.

Then to find dx I set arctany=x and solved to find dx=dy/(1=y^2) Then I substituted my y values back into my original integral so

. . . . . . . .integral tan^3xdx

. . . . . . . .=integral y^3dy/(y^2+1)

I solved this integral ( after dividing to get integral ydy minus integral ydy/(y^2+1) ) and found that the integral was

. . . . . . . .(y^2)/2-lnabs(y^2+1)+C

Then I substituted x back into the solution for y (using the substitution y=tanx My final answer was

. . . . . . . .(tan^2x)/2-lnabs(tan^2x+1)/2+C

Could you please tell me if this method of solving the integral is correct? Also, what is the better way to solve the integral? Thank you!

 

[dora]

Integral of tan ^ 3 x

Hello Marian 51,

I think your approach in solving this problem is correct. But you can also use trigonometric substitution to solve the problem:

tan³x = ( tan² x ) tan x

. . . . . . . .= ( sec² x - 1 ) tan x

. . . . . . . .= sec² x tan x - tan x

. . . . . . . .= sec x sec x tan x - tan x

Let I = integral ( sec x sec x tan x - tan x ) d x. Splitting the subtraction into two integrals, we get I = I₁ - I₂. Working separately, we get:

I₁ = integral ( secx sec x tanx ) dx

Let sec(x) = y, so sec(x) tan(x) dx = dy. Then:

I₁ = integral y dy

. . . . . . . .= y² / 2 + c₁

. . . . . . . .= ( sec²x ) + c₁

I₂ = integral tan x dx

. . . . . . . .= ln | sec x | + c₂

I = I₁ - I₂

. . . . . . . .= (sec²x) / 2 +c₁ - ln| sec x | + c₂

. . . . . . . .= ( sec²x ) - ln | cos x |^(-1) + C

. . . . . . . .= sec²x + ln| cos x | + C

...where c₁ + c₂ = C.

Is this clear? If not, please reply with your questions. Thank you!

 

[soroban]

Re: Integrate tan^3x dx

Hello, marian51!

Yes, a little trig manipulation is in order . . .

\(\displaystyle \L\int\tan^3x\,dx\)

We have: \(\displaystyle \:\tan^3x\:=\:\tan^2x\cdot\tan x \:=\\sec^2x\,-\,1)\cdot\tan x\:=\:\sec^2x\cdot\tan x\,-\,tan x\)

The integral becomes: \(\displaystyle \L\int\tan x\cdot\sec^2x\,dx \,-\,\int\tan x\,dx\)

. . In the first integral, let \(\displaystyle u\,=\,tan x\)

And we have: \(\displaystyle \L\:\frac{1}{2}\tan^2x\,+\,\ln|\cos x|\,+\,C\)

 

[marian51]

Thank you Dora and Soroban!