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Question Number 109342 by 150505R last updated on 22/Aug/20

Commented by maths mind last updated on 23/Aug/20

$$x=tg\left(t\right),dx=(1+{tg}^2\left(t\right))dt$$$$={\int }_0^{\frac{\pi }{4}}ln\left(\frac{{cos}^2\left(t\right)-{sin}^2\left(t\right)}{sin\left(t\right)cos\left(t\right)}\right).dt$$$$={\int }_0^{\frac{\pi }{2}}ln\left(2cot\left(r\right)\right).\frac{dr}{2}$$$$=\frac{\pi }{4}ln\left(2\right)+{\int }_0^{\frac{\pi }{2}}ln\left(cos\left(r\right)\right)dr-{\int }_0^{\frac{\pi }{2}}ln\left(sin\left(r\right)\right)dr$$$$=\frac{\pi ln\left(2\right)}{4}$$$$$$

Commented by mathdave last updated on 23/Aug/20

$$youshouldveexplainbetterhowyouchange$$$$\frac{{\cos }^{2}t-{\sin }^{2}t}{\sin t\cos t}=2\cot \left(r\right)forthosethatarenotpro$$$$likeu.$$$$letmeputlightonthatarea$$$$$$

Commented by mathdave last updated on 23/Aug/20

$$weknownthat\cos 2t={\cos }^{2}t-{\sin }^{2}tand$$$$\frac{\sin 2t}{2}=\cos t\sin tso\frac{{cos}^{2}t-{sin}^{2}t}{sintcost}=2\frac{cos2t}{sin2t}=2cot\left(2t\right)$$$$letr=2tanddt=\frac{dt}{2}$$$$atlimitt\Rightarrow \frac{\pi }{4},r\Rightarrow \frac{\pi }{2}andatlimitt\Rightarrow 0,r\Rightarrow 0$$$$note{\int }_0^{\frac{\pi }{2}}\ln \left(\cos r\right)={\int }_0^{\frac{\pi }{2}}\ln \left(\sin x\right)=-\frac{\pi }{2}\ln 2$$

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