Integrate {eq}tan (\frac{ln x}{x}) dx= {/eq}

Question:

Integrate {eq}tan (\frac{ln x}{x}) dx= {/eq}

Integration:

The indefinite integral does not contain the limit of integration that is known as the lower limit and the upper limits. If the function is {eq}f\left( x \right)dx {/eq} is solved by using the indefinite integral it gives the solution as {eq}\int {f\left( x \right)dx = F\left( x \right) + C} {/eq} , where {eq}C {/eq} is the constant.

Given data

• The given expression is {eq}\int {\tan } \left( {\dfrac{{\ln x}}{x}} \right)dx {/eq}.

Suppose {eq}u = \ln x {/eq}.

Differentiate the above equation,

{eq}du = \dfrac{1}{x}dx {/eq}

Therefore, from given expression,

{eq}\begin{align*} \int {\tan } \left( {\dfrac{{\ln x}}{x}} \right)dx &= \int {\tan udu} \\ & = \int {\dfrac{{\sin u}}{{\cos u}}du.....\left( 1 \right)} \end{align*} {/eq}

Now, suppose {eq}\cos u = t {/eq}.

Differentiate the above equation,

{eq}dt = - \sin udu {/eq}

So, from equation (1),

{eq}\begin{align*} \int {\tan } \left( {\dfrac{{\ln x}}{x}} \right)dx& = \int { - \dfrac{1}{t}} dt\\ & = - \ln \left| t \right|....\left( 2 \right) \end{align*} {/eq}

Substitute {eq}\cos \left( {\ln x} \right) {/eq} for {eq}t {/eq} in equation (2).

{eq}\int {\tan } \left( {\dfrac{{\ln x}}{x}} \right)dx = - \ln \left| {\cos \left( {\ln x} \right)} \right| + C {/eq}

Thus, the required solution is {eq}- \ln \left| {\cos \left( {\ln x} \right)} \right| + C {/eq}.