Question

integrate sin^2 into log x by X​

Answer 1

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

$\rm : \implies \dfrac{1}{2} logx - \dfrac{1}{4} \sin2 logx + c$