CBSE Class 12 Math 2009 Solved Paper

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Question : 25 of 29
 
Marks: +1, -0
Evaluate: ∫0πecosxecosx+e−cosx dx
OR
Evaluate: ∫0π2 (2 log sin x - log sin 2x) dx
Solution:
Let I = ∫0πecosxecosx+e−cosx dx
Using ∫0a f (x) = ∫0a f (a - x) dx
I = ∫0πecos(π−x)ecos(π−x)+e−cos(π−x) dx
2I = ∫0π e−cosx+ecosxecosx+e−cosx dx
I = 12∫0π dx = 12 [π - 0] = π2
OR
I = ∫0π2 (2 log sin x - log sin 2x) dx
I = ∫0π2 (logsin2x2sinx.cosx.dx)
I = ∫0π2 log (tanx2) . dx ... (i)
Using property ∫0a f (x) dx = ∫0a f (a - x) dx
We get,
I = ∫0π2 log (tan(π2−x)2) dx
⇒ I = ∫0π2 log (cotx2) dx ... (ii)
Additing (i)&(ii)
2I = ∫0π2 [log(tanx2)+log(cotx2)] dx
⇒ 2I = ∫0π2 log [(tanx2)(cotx2)] dx
⇒ I = 12∫0π2 log (14) dx
⇒ I = 12 log (14)×(π2)
⇒ I = 12 log (14)12 × (π2)
⇒ I = log (12)×(π2)
⇒ I = π2log12
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