Integrate sin^2 into log x by X

1.	Integrate sin^2 into log x by X
Answer» Solution:- $: \implies\rm \int \dfrac{ \sin^{2} log x }{x} \: dx \\$ :- USING substitution method $\rm \: let$ $\rm\to \: logx = t$ :- differentiate on both side $\rm \to \dfrac{1}{x} dx= dt$ Now take $: \implies\rm \int \dfrac{ \sin^{2} log x }{x} \: dx \\$ $\rm : \implies \int { \sin {}^{2} t} \: dt \\$ Now we can write $\rm \to1 - 2 \sin {}^{2} t = \cos2t$ $\rm \to \: 1 - \cos2t = 2 \sin {}^{2} t$ $\rm \to \sin {}^{2} t = \dfrac{1 - \cos2t}{2}$ Now substitute the value of sin²t $\rm : \implies \int \dfrac{1 - \cos2t }{2} dt \\$ Take 1/2 out side the integration because it is constant TERM $\rm : \implies \dfrac{1}{2} \int1 - \cos2t \: dt \\$ By integration we get $\rm : \implies \: \dfrac{1}{2} \bigg(t - \dfrac{1}{2} \sin2t \bigg) + c$ $\rm : \implies \dfrac{1}{2} t - \dfrac{1}{4} \sin2t + c$ Now PUT the value of t $\rm : \implies \dfrac{1}{2} logx - \dfrac{1}{4} \sin2 logx + c$ ANSWER $\rm : \implies \dfrac{1}{2} logx - \dfrac{1}{4} \sin2 logx + c$

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