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Question Number 91910 by Ar Brandon last updated on 03/May/20

$$\int \frac{\ln (1+{\sin }^2\mathrm{x})}{{\sin }^2\mathrm{x}}\mathrm{dx}$$

Commented by mathmax by abdo last updated on 03/May/20

$$letf\left(a\right)=\int \frac{ln(1+{asin}^{2}x)}{{sin}^{2}x}witha>0$$$$f^{{}^{\prime }}\left(a\right)=\int ln(1+{asin}^{2}x)dx=\int ln(1+a\times \frac{1-cos\left(2x\right)}{2})dx$$$$=\int ln(2+a-acos\left(2x\right))dx-xln\left(2\right)$$$$=\int ln((2+a)(1-\frac{a}{2+a}cos\left(2x\right))dx-xln\left(2\right)$$$$=(ln(2+a)-ln2)x+\int ln(1-\frac{a}{a+2}cos\left(2x\right))letdetermine$$$$I_{\lambda }=\int ln(1-\lambda cos\left(2x\right))dxwith0<\lambda <1$$$$\frac{{dI}_{\lambda }}{d\lambda }=-\int \frac{cos\left(2x\right)}{1-\lambda cos\left(2x\right)}dx{=}_{2x=t}\frac{1}{2}\int \frac{cos\left(t\right)}{1-\lambda cost}dt$$$$=-\frac{1}{2\lambda }\int \frac{1-\lambda cost-1}{1-\lambda cost}dt=-\frac{1}{2\lambda }t+\frac{1}{2\lambda }\int \frac{dt}{1-\lambda cost}$$$$\int \frac{dt}{1-\lambda cost}{=}_{tan\left(\frac{x}{2}\right)=u}\int \frac{2du}{(1+u^2)(1-\lambda \frac{1-u^2}{1+u^2})}$$$$=2\int \frac{du}{1+u^2-\lambda +\lambda u^2}=2\int \frac{du}{(1+\lambda )u^2+1-\lambda }$$$$=\frac{2}{1+\lambda }\int \frac{du}{u^2+\frac{1-\lambda }{1+\lambda }}{=}_{u=\sqrt{\frac{1-\lambda }{1+\lambda }}z}\frac{2}{1+\lambda }\int \frac{1}{\frac{1-\lambda }{1+\lambda }(1+z^2)}\times \frac{\sqrt{1-\lambda }}{\sqrt{1+\lambda }}dz$$$$=\frac{2}{\sqrt{1-{\lambda }^2}}\times \frac{\pi }{2}=\frac{\pi }{\sqrt{1-{\lambda }^2}}\Rightarrow I{}_{\lambda }^{\prime }=-\frac{x}{\lambda }+\frac{\pi }{2\lambda \sqrt{1-{\lambda }^2}}\Rightarrow$$$$I_{\lambda }=-xln\lambda +\frac{\pi }{2}\int \frac{d\lambda }{\lambda \sqrt{1-{\lambda }^2}}+C$$$$\int \frac{d\lambda }{\lambda \sqrt{1-{\lambda }^2}}{=}_{\lambda =sint}\int \frac{costdt}{sintcost}=\int \frac{dt}{sint}{=}_{tan\left(\frac{t}{2}\right)=u}$$$$=\int \frac{2du}{(1+u^2)\frac{2u}{1+u^2}}=\int \frac{du}{u}=ln\mid u\mid =ln\mid tan\left(\frac{arcsin\lambda }{2}\right)\mid \Rightarrow$$$$I_{\lambda }=-xln\lambda +\frac{\pi }{2}ln\mid tan\left(\frac{arcsin\lambda }{2}\right)\mid +C\Rightarrow$$$$f^{{}^{\prime }}\left(a\right)=xln\left(\frac{2+a}{2}\right)-xln\left(\frac{a}{a+2}\right)+\frac{\pi }{2}ln\mid tan\left(\frac{arcsin\left(\frac{a}{a+2}\right)}{2}\right)\mid +C$$$$\Rightarrow f\left(a\right)=x\int (ln\left(\frac{2+a}{2}\right)-ln\left(\frac{a}{a+2}\right)\mid +\frac{\pi }{2}\int ln\mid tan\left(\frac{arcsin\left(\frac{a}{a+2}\right)}{2}\right)+C$$$$....becontinued....$$

Commented by Ar Brandon last updated on 03/May/20

�� thanks

Commented by mathmax by abdo last updated on 03/May/20

$$youarewelcome$$

Commented by Ar Brandon last updated on 03/May/20

I got this. What do think ? ����

Commented by turbo msup by abdo last updated on 03/May/20

$$thisisthewaytheproblemehere$$$$isthstintegrsliswithoutlimits...$$

Answered by Ar Brandon last updated on 03/May/20

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