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Question Number 201980 by Calculusboy last updated on 17/Dec/23

Answered by Sutrisno last updated on 18/Dec/23

misal x^2 =t dx=(dt/(2x)) =∫x.x^2 cot(x^2 )(dt/(2x)) =(1/2)∫tcot(t)dt =(1/2)(t.ln∣sint∣−∫ln∣sint∣dt) (1) I=∫ln∣sint∣dt I=∫ln∣2.sin(1/2)t.cos(1/2)t∣dt I=∫ln2 dt +∫∣sin(1/2)t.cos(1/2)t∣dt misal t=π−y → −dt=dy I=∫ln2 dt +∫∣sin(1/2)(π−y).cos(1/2)(π−y)∣.−dy I=∫ln2 dt −∫∣cos(1/2)(y).sin(1/2)(y)∣dy I=∫ln2 dt −(1/2)∫∣sin(y)∣dy I=∫ln2 dt −(1/2)I (3/2)I=tln2 → I=(2/3)tln2 =(1/2)(t.ln∣sint∣−(2/3)t.ln2)+c =(1/2)(x^2 .ln∣sinx^2 ∣−(2/3)x^2 .ln2)+c

$${misal}\:{x}^{\mathrm{2}} ={t} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{dx}=\frac{{dt}}{\mathrm{2}{x}} \\ $$$$=\int{x}.{x}^{\mathrm{2}} {cot}\left({x}^{\mathrm{2}} \right)\frac{{dt}}{\mathrm{2}{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int{tcot}\left({t}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({t}.{ln}\mid{sint}\mid−\int{ln}\mid{sint}\mid{dt}\right)\:\:\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$${I}=\int{ln}\mid{sint}\mid{dt} \\ $$$${I}=\int{ln}\mid\mathrm{2}.{sin}\frac{\mathrm{1}}{\mathrm{2}}{t}.{cos}\frac{\mathrm{1}}{\mathrm{2}}{t}\mid{dt} \\ $$$${I}=\int{ln}\mathrm{2}\:{dt}\:+\int\mid{sin}\frac{\mathrm{1}}{\mathrm{2}}{t}.{cos}\frac{\mathrm{1}}{\mathrm{2}}{t}\mid{dt} \\ $$$${misal}\:{t}=\pi−{y}\:\rightarrow\:−{dt}={dy} \\ $$$${I}=\int{ln}\mathrm{2}\:{dt}\:+\int\mid{sin}\frac{\mathrm{1}}{\mathrm{2}}\left(\pi−{y}\right).{cos}\frac{\mathrm{1}}{\mathrm{2}}\left(\pi−{y}\right)\mid.−{dy} \\ $$$${I}=\int{ln}\mathrm{2}\:{dt}\:−\int\mid{cos}\frac{\mathrm{1}}{\mathrm{2}}\left({y}\right).{sin}\frac{\mathrm{1}}{\mathrm{2}}\left({y}\right)\mid{dy} \\ $$$${I}=\int{ln}\mathrm{2}\:{dt}\:−\frac{\mathrm{1}}{\mathrm{2}}\int\mid{sin}\left({y}\right)\mid{dy} \\ $$$${I}=\int{ln}\mathrm{2}\:{dt}\:−\frac{\mathrm{1}}{\mathrm{2}}{I} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}{I}={tln}\mathrm{2}\:\rightarrow\:{I}=\frac{\mathrm{2}}{\mathrm{3}}{tln}\mathrm{2} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({t}.{ln}\mid{sint}\mid−\frac{\mathrm{2}}{\mathrm{3}}{t}.{ln}\mathrm{2}\right)+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} .{ln}\mid{sinx}^{\mathrm{2}} \mid−\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{2}} .{ln}\mathrm{2}\right)+{c} \\ $$$$ \\ $$$$ \\ $$

Commented by Calculusboy last updated on 18/Dec/23

thanks sir

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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