Definite integral: ∫(tan^3(x)+tan(x))dx = ∫tan(x)(tan^2(x)+1)dx = ∫ tan(x) (1/cos^2(x

[Anonymous11]

Definite integral: ∫(tan^3(x)+tan(x))dx = ∫tan(x)(tan^2(x)+1)dx = ∫ tan(x) (1/cos^2(x

∫(tan^3(x)+tan(x))dx = ∫tan(x)(tan^2(x)+1)dx = ∫ tan(x) (1/cos^2(x) ) dx
And from there I say t=tan(x) => dt= 1/cos^2(x) dx and then ∫ t dt = t^2/2+c which yields tan^2(x)/2+c
But if I solve this way : ∫ tan(x) (1/cos^2(x) ) dx =∫ sin(x)/cos^3(x) dx
and then t=cos(x) => dt=-sin(x) dx
so it equals -∫ 1/t^3 dt =1/(2t^2) +c so substituing back it equals 1/(2cos^2(x))+c
What troubles me is that the two results are not equal .
Is there anything wrong ?

 

[Dr.Peterson]

∫(tan^3(x)+tan(x))dx = ∫tan(x)(tan^2(x)+1)dx = ∫ tan(x) (1/cos^2(x) ) dx
And from there I say t=tan(x) => dt= 1/cos^2(x) dx and then ∫ t dt = t^2/2+c which yields tan^2(x)/2+c
But if I solve this way : ∫ tan(x) (1/cos^2(x) ) dx =∫ sin(x)/cos^3(x) dx
and then t=cos(x) => dt=-sin(x) dx
so it equals -∫ 1/t^3 dt =1/(2t^2) +c so substituing back it equals 1/(2cos^2(x))+c
What troubles me is that the two results are not equal .
Is there anything wrong ?

No, nothing is wrong. You just have different values for C!

Your second answer is sec^2(x)/2 + C, which by the identity sec^2(a) = 1 + tan^2(a), is equal to (1 + tan^2(x))/2 + C = tan^2(x)/2 + (C + 1/2). So this is identical to the first answer, with the C there replaced by C + 1/2.

This is a classic "non-error" in integration; I see it all the time. You get a valid answer, but think you are wrong because it differs from the book's answer, or your neighbor's.

 

[Steven G]

∫(tan^3(x)+tan(x))dx = ∫tan(x)(tan^2(x)+1)dx = ∫ tan(x) (1/cos^2(x) ) dx
And from there I say t=tan(x) => dt= 1/cos^2(x) dx and then ∫ t dt = t^2/2+c which yields tan^2(x)/2+c
But if I solve this way : ∫ tan(x) (1/cos^2(x) ) dx =∫ sin(x)/cos^3(x) dx
and then t=cos(x) => dt=-sin(x) dx
so it equals -∫ 1/t^3 dt =1/(2t^2) +c so substituing back it equals 1/(2cos^2(x))+c
What troubles me is that the two results are not equal .
Is there anything wrong ?

In these cases I would compute tan^2(x)/2 - 1/(2cos^2(x) for a few x-values and see if I get the same results. If I do, then it seems that the two results differ by a constant which is ok. Then try to prove (ie show) that the two results do differ by a constant (as Dr P did)