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Question
Wed February 02, 2011

# Integrate the given ques

Wed March 23, 2011
Dear student,

sinx + cosx = â2.sin(x+â/4)

So, log(sinx + cosx) = log[â2.sin(x+â/4)]
= logâ2  + logsin(x+â/4)

We have:
I = â«logâ2 dx + â«logsin(x+â/4) dx
= (logâ2)x + I'

In I', put x+â/4 = t. So, dx = dt

I' = â«log(sint) dt

Now, the limit is 0 to â/2

So, I = â/2.logâ2 + I', where the limits are from o to â/2  ... (A)

Now, I'=â« log(sin x).dx ...(1)
I'=â«logsin(Ï/2-x)dx
I'=â«logcosx dx...(2)
2I'=â«logsinx+logcosx dx
=â«logsinx*cosx dx
=â«log(sin2x)/2 dx
=â«[logsin2x-log2] dx
=â«logsin2xdx - â«log2dx ...(3)
In the first integral put 2x=t so dx=dt/2
when x=0,t=0 when x=Ï/2 ,t=Ï
So, the first integral can be solved as below:
â«logsint dt with limit t= o & t=Ï
Now, solve this integral and get the value of the integral I' from the equation (3). Thereafter, equate the value of I' in (A) to get the value of the required integral I.

We hope that clarifies your query.

Regards,
Team
TopperLearning
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