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

Answered by Sutrisno last updated on 18/Dec/23

$$misal{x}^2=t$$$$dx=\frac{dt}{2x}$$$$=\int x.x^{2}cot\left(x^2\right)\frac{dt}{2x}$$$$=\frac{1}{2}\int tcot\left(t\right)dt$$$$=\frac{1}{2}(t.ln\mid sint\mid -\int ln\mid sint\mid dt)\left(1\right)$$$$$$$$I=\int ln\mid sint\mid dt$$$$I=\int ln\mid 2.sin\frac{1}{2}t.cos\frac{1}{2}t\mid dt$$$$I=\int ln2dt+\int \mid sin\frac{1}{2}t.cos\frac{1}{2}t\mid dt$$$$misalt=\pi -y\to -dt=dy$$$$I=\int ln2dt+\int \mid sin\frac{1}{2}(\pi -y).cos\frac{1}{2}(\pi -y)\mid .-dy$$$$I=\int ln2dt-\int \mid cos\frac{1}{2}\left(y\right).sin\frac{1}{2}\left(y\right)\mid dy$$$$I=\int ln2dt-\frac{1}{2}\int \mid sin\left(y\right)\mid dy$$$$I=\int ln2dt-\frac{1}{2}I$$$$\frac{3}{2}I=tln2\to I=\frac{2}{3}tln2$$$$=\frac{1}{2}(t.ln\mid sint\mid -\frac{2}{3}t.ln2)+c$$$$=\frac{1}{2}(x^2.ln\mid {sinx}^2\mid -\frac{2}{3}{x}^2.ln2)+c$$$$$$$$$$

Commented by Calculusboy last updated on 18/Dec/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

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