Question

5. (a) Integrate
∫ sin(a+b logx)/x dx​

Answer 1

Given Question :-

$\rm : \implies \: \int \: \dfrac{sin \: (a \: + \: b \: logx) }{x} \: dx$

Solution

Method used

• Method of substitution

Formula used

$1. \: \boxed{ \pink{ \rm \:\dfrac{d}{dx} log(x) = \dfrac{1}{x} }}$

$2. \: \boxed{ \pink{ \rm \: \int \: sinx \: dx \: = \: - \: cosx \: + \: c }}$

Now, Consider

$\rm : \implies \: Let \: I \: = \: \int \: \dfrac{sin \: (a \: + \: b \: logx) }{x} \: dx$

$\rm : \implies \: Put \: a \: + \: b \: logx \: = \: y$

On differentiating both sides w. r. t. x, we get

$\rm : \implies \: \dfrac{b}{x} = \dfrac{dy}{dx}$

$\rm : \implies \: \dfrac{1}{x} dx \: = \: \dfrac{dy}{b}$

Hence, given integral can be rewritten as

$\rm : \implies \: I \: = \int \: sin \: y \: \dfrac{dy}{b}$

$\rm : \implies \: \dfrac{1}{b} \int \: siny \: dy$

$\rm : \implies \: - \dfrac{1}{b} \: cos \: y \: + \: c$

$\rm : \implies \: - \dfrac{1}{b} \: cos \: (a\: + \: b \: logx) \: + \: c$