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Question Number 175471 by mnjuly1970 last updated on 31/Aug/22

Answered by Ar Brandon last updated on 31/Aug/22

$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{2}}\sin 2x{\ln }^2(1+\cos x)dx=2{\int }_0^{\frac{\pi }{2}}\sin x\cos x{\ln }^2(1+\cos x)dx$$$$=2{\int }_0^{1}t{\ln }^2(1+t)dt={\left[t^2{\ln }^2(1+t)\right]}_0^1-2{\int }_0^1\frac{t^2\ln (1+t)}{1+t}dt$$$$={\ln }^2\left(2\right)-2{\int }_0^1(t-1+\frac{1}{1+t})\ln (1+t)dt$$$$={\ln }^2\left(2\right)-2{\int }_0^1(t-1)\ln (1+t)dt-2{\int }_0^1\frac{\ln (1+t)}{1+t}dt$$$$={\ln }^2\left(2\right)-2{\left[(\frac{t^2}{2}-t)\ln (1+t)\right]}_0^1+{\int }_0^1\frac{t^2-2t}{t+1}dt-{\left[{\ln }^2(1+t)\right]}_0^1$$$$={\ln }^2\left(2\right)+\ln 2+{\int }_0^1(t-3+\frac{3}{t+1})dt-{\ln }^2\left(2\right)$$$$=\ln 2+{[\frac{t^2}{2}-3t+3\ln (t+1)]}_0^1=\ln 2-\frac{5}{2}+3\ln 2$$$$=\frac{8\ln 2-5}{2}=\frac{2^3\ln 2-5}{2},\Rightarrow p+q=2+3=5$$

Commented by mnjuly1970 last updated on 31/Aug/22

$$thanksalotsir...$$

Commented by Ar Brandon last updated on 31/Aug/22

You're welcome, Sir !

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