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Question Number 142643 by ArielVyny last updated on 03/Jun/21

$${\int }_0^{\frac{\pi }{2}}\frac{{cos}^{2}t}{sint}dt$$

Answered by MJS_new last updated on 03/Jun/21

$$\int \frac{{\cos }^{2}t}{\sin t}dt=\int \frac{1-{\sin }^{2}t}{\sin t}dt=\int (-\sin t+\csc t)dt=$$$$=\cos t+\ln \tan \frac{t}{2}+C$$$$\Rightarrow \mathrm{Integral}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{converge}$$

Commented by ArielVyny last updated on 03/Jun/21

$$in[0.\frac{\pi }{2}]integraldoesnotconverge?$$

Commented by MJS_new last updated on 03/Jun/21

$$\mathrm{yes}.$$$$\text{Missing left or extra right}$$$${\left[\ln \tan \frac{x}{2}\right]}_0^{\pi /2}=0-\underset{x\to 0^{+}}{\lim }\ln \tan \frac{x}{2}\mathrm{but}\mathrm{this}\mathrm{limit}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$

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