Question

Find ∫ sin⁡x log⁡(cos⁡x) dx.

(a) cos⁡x (log⁡(sin⁡x)-1)+C

(b) sin⁡x (log⁡(cos⁡x)+1)+C

(c) cos⁡x (log⁡(cos⁡x)-1)+C

(d) cos⁡x (log⁡(cos⁡x)-1)+C

The question was asked in final exam.

My question is taken from Integration by Parts in section Integrals of Mathematics – Class 12

Answer 1

Right choice is (c) cos⁡x (log⁡(cos⁡x)-1)+C

To explain I would say: Let cos⁡x=t

Differentiating w.r.t x, we get

-sin⁡x dx=dt

sin⁡x dx=-dt

∴∫ sin⁡x log⁡(cos⁡x) dx=∫ -log⁡t dt

Using ∫ u.v dx=u∫ v dx-∫ u’ (∫ v dx) , we get

∫ -log⁡t dt=-log⁡t ∫ 1 dt-∫ (-log⁡t)’ ∫ 1 dt

=-t log⁡t+∫ dt

=-t log⁡t+t

=t(log⁡t-1)

Replacing t with cos⁡x, we get

∫ sin⁡x log⁡(cos⁡x) dx=cos⁡x (log⁡(cos⁡x)-1)+C