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Question: Find the Indefinite integral of \[\int {\cos 2\theta \log \left( {\dfrac{{\cos \theta + \sin \theta ...

Find the Indefinite integral of cos2θlog(cosθ+sinθcosθsinθ)dθ\int {\cos 2\theta \log \left( {\dfrac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}} \right)} d\theta

Explanation

Solution

Hint : In mathematics, an integral assigns a number of functions in a way that can describe displacement area, volume, and other concepts that arise by combining infinitesimal sections. Its inverse operation is differentiation. The given question, since the integrals have no upper and lower limit, hence we say the integral is an indefinite integral, and this is also called Antiderivative. The indefinite integral is not just a function; it is the family of the function.
In the question, we can see the given function is the product of two functions hence we use the uv rule to integrate this function given as uvdx=uvdxu(vdx)dx\int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)} } } dx

Complete step-by-step answer :
Given function I=cos2θlog(cosθ+sinθcosθsinθ)dθI = \int {\cos 2\theta \log \left( {\dfrac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}} \right)} d\theta
We can write the function as
I=(cos2θsin2θ)log(cosθ+sinθcosθsinθ)dθ\Rightarrow I = \int {\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)\log \left( {\dfrac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}} \right)} d\theta
By using the trigonometric identities cos2x=cos2xsin2x\cos 2x = {\cos ^2}x - {\sin ^2}x
Now by using the logarithm quotient rule logxy=logxlogy\log \dfrac{x}{y} = \log x - \log y, we can further write
I=(cos2θsin2θ)[log(cosθ+sinθ)log(cosθsinθ)]dθ\Rightarrow I = \int {\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)\left[ {\log \left( {\cos \theta + \sin \theta } \right) - \log \left( {\cos \theta - \sin \theta } \right)} \right]} d\theta
By further solving we can write
I=[(cosθ+sinθ)(cosθsinθ)log(cosθ+sinθ)]dθ[(cosθ+sinθ)(cosθsinθ)log(cosθsinθ)]dθ(i)\Rightarrow I = \int {\left[ {\left( {\cos \theta + \sin \theta } \right)\left( {\cos \theta - \sin \theta } \right)\log \left( {\cos \theta + \sin \theta } \right)} \right]d\theta - \int {\left[ {\left( {\cos \theta + \sin \theta } \right)\left( {\cos \theta - \sin \theta } \right)\log \left( {\cos \theta - \sin \theta } \right)} \right]} } d\theta - - - - (i)
Let cosθ+sinθ=u\cos \theta + \sin \theta = u
By differentiating with respect to θ\theta we get

(sinθ+cosθ)dθ=du (cosθsinθ)dθ=du \Rightarrow \left( { - \sin \theta + \cos \theta } \right)d\theta = du \\\ \Rightarrow \left( {\cos \theta - \sin \theta } \right)d\theta = du \\\

Also, let cosθsinθ=t\cos \theta - \sin \theta = t
By differentiating with respect to θ\theta we get

(sinθcosθ)dθ=dt (sinθ+cosθ)dθ=dt \Rightarrow \left( { - \sin \theta - \cos \theta } \right)d\theta = dt \\\ \Rightarrow - \left( {\sin \theta + \cos \theta } \right)d\theta = dt \\\

By substituting the above two equations, in the equation (i) we get

I=[(cosθ+sinθ)(cosθsinθ)log(cosθ+sinθ)]dθ[(cosθ+sinθ)(cosθsinθ)log(cosθsinθ)]dθ I=ulogudu+tlogtdt(ii) \Rightarrow I = \int {\left[ {\left( {\cos \theta + \sin \theta } \right)\left( {\cos \theta - \sin \theta } \right)\log \left( {\cos \theta + \sin \theta } \right)} \right]d\theta - \int {\left[ {\left( {\cos \theta + \sin \theta } \right)\left( {\cos \theta - \sin \theta } \right)\log \left( {\cos \theta - \sin \theta } \right)} \right]} } d\theta \\\ \Rightarrow I = \int {u\log udu + \int {t\log tdt} } - - - - (ii) \\\

Now use uv rule for further operation in both terms of the equation (ii) we get,

I=loguudud(logu)du(udu)du+logttdtd(logt)dt(tdt)dt =[logu.u221u.u22du]+[logt.t221t.t22du] \Rightarrow I = \log u\int {udu} - \int {\dfrac{{d(\log u)}}{{du}}\left( {\int {udu} } \right)} du + \log t\int {tdt} - \int {\dfrac{{d(\log t)}}{{dt}}\left( {\int {tdt} } \right)} dt \\\ = \left[ {\log u.\dfrac{{{u^2}}}{2} - \int {\dfrac{1}{u}.\dfrac{{{u^2}}}{2}du} } \right] + \left[ {\log t.\dfrac{{{t^2}}}{2} - \int {\dfrac{1}{t}.\dfrac{{{t^2}}}{2}du} } \right] \\\

This is equal to
I=[logu.u22u2du]+[logt.t22t2du]\Rightarrow I = \left[ {\log u.\dfrac{{{u^2}}}{2} - \int {\dfrac{u}{2}du} } \right] + \left[ {\log t.\dfrac{{{t^2}}}{2} - \int {\dfrac{t}{2}du} } \right]
By further solving
I=[logu.u22u24]+[logt.t22t24]\Rightarrow I = \left[ {\log u.\dfrac{{{u^2}}}{2} - \dfrac{{{u^2}}}{4}} \right] + \left[ {\log t.\dfrac{{{t^2}}}{2} - \dfrac{{{t^2}}}{4}} \right]
Now substituting the values of u and t in the obtained function

I=[logu.u22u24]+[logt.t22t24] =[(cosθ+sinθ)22log(cosθ+sinθ)(cosθ+sinθ)24]+[(cosθsinθ)22log(cosθsinθ)(cosθsinθ)24] \Rightarrow I = \left[ {\log u.\dfrac{{{u^2}}}{2} - \dfrac{{{u^2}}}{4}} \right] + \left[ {\log t.\dfrac{{{t^2}}}{2} - \dfrac{{{t^2}}}{4}} \right] \\\ = \left[ {\dfrac{{{{\left( {\cos \theta + \sin \theta } \right)}^2}}}{2}\log \left( {\cos \theta + \sin \theta } \right) - \dfrac{{{{\left( {\cos \theta + \sin \theta } \right)}^2}}}{4}} \right] + \left[ {\dfrac{{{{\left( {\cos \theta - \sin \theta } \right)}^2}}}{2}\log \left( {\cos \theta - \sin \theta } \right) - \dfrac{{{{\left( {\cos \theta - \sin \theta } \right)}^2}}}{4}} \right] \\\

Important functions used:
Trigonometric identities: cos2x=cos2xsin2x\cos 2x = {\cos ^2}x - {\sin ^2}x
Quotient rule logxy=logxlogy\log \dfrac{x}{y} = \log x - \log y

Note : For better understanding, students can say that integration is a way of adding slices to find the whole. Integration can be used to find areas, volumes, central points, and many useful things. Moreover, students should be aware before starting which term in the expression should be taken as u and which for v.