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How do I find the integral #intcos(x)ln(sin(x))dx# ?

Answer 1

Charles Anctil

#=ln(sin(x))(sin(x)-1)+c#, where #c# is a constant

Explanation

#=intcos(x)ln(sin(x))dx#

let's #sin(x)=t#, #=># #cos(x)dx=dt#

#=intln(t)*1dt#

Using Integration by Parts,

#int(I)(II)dx=(I)int(II)dx-int((I)'int(II)dx)dx#

where #(I)# and #(II)# are functions of #x#, and #(I)# represents which will be differentiated and #(II)# will be integrated subsequently in the above formula

Similarly following for the problem,

#=ln(t)int1*dt-int((lnt)'intdt)dt#

#=tln(t)-int1/t*tdt#

#=tln(t)-t+c#, where #c# is a constant

#=t(ln(t)-1)+c#, where #c# is a constant

#=sin(x)(lnsin(x)-1)+c#, where #c# is a constant

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Answer 2

Ezekiel Alexander

To find the integral of ∫cos(x)ln(sin(x))dx, you can use integration by parts. Let u = ln(sin(x)) and dv = cos(x)dx. Then, differentiate u to find du, and integrate dv to find v. Afterward, apply the integration by parts formula:

∫udv = uv - ∫vdu

Finally, substitute the expressions for u, dv, v, and du into the formula and evaluate the integral.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.