Question

Integrate 3 sec^2⁡x log⁡(tan⁡x) dx.

(a) -log⁡(tan⁡x) (tan⁡x-1)+C

(b) log⁡(tan⁡x) (sec⁡x+1)+C

(c) tan⁡x (log⁡(tan⁡x)-1)+C

(d) tan⁡x (log⁡sec⁡x +1)+C

I had been asked this question in my homework.

My question is based upon Integration by Parts topic in division Integrals of Mathematics – Class 12

Answer 1

Right answer is (c) tan⁡x (log⁡(tan⁡x)-1)+C

Easiest explanation: By using∫ u.v dx=u∫ v dx-∫ u'(∫ v dx), we get

∫ log⁡(tan⁡x) sec^2x dx=log⁡(tan⁡x) ∫ sec^2 x⁡dx -∫ (log⁡tan⁡x)’∫ sec^2x dx

=tan⁡x log⁡(tan⁡x)-\(\int \frac{1}{tan⁡x} sec^2⁡x.tan⁡x \,dx\)

=tan x⁡log⁡(tan⁡x)-∫ sec^2⁡x dx

=tan x⁡log⁡(tan⁡x)-tan⁡x+C

=tan⁡x (log⁡(tan⁡x)-1)+C