Question

Solve (xiii) and (xvi)


INTEGRALS.

Answer 1

Hello,

Solution:

The integration of above mentioned functions can be easily done by substitution.

1) In first solution, put log x = u.

2) differentiate log x = u, with respect to x, find the value of dx in terms of du.

3) After substitution,integrate the expression, undo substitution.

by the same way another integration can be done.

please find the solution in attachment.

Hope it helps you.

Answer 2

$(xiii.) \: \: I \: = \int{} \: \dfrac{ \sin(log \: x) \: dx} {x} \\ \\ Let \: \: l og \: x \: \: = \: \: t \\ \: \: \dfrac{1}{x} \: dx \: = \: dt \\ \\ \: Substituting \: in \: I \: , \\ \\ \int{} sin \: t \: dt \: = - \: cos \: t \: + \: C \\ \\ \large \boxed{ I \: \: = \: \: - \: cos \: ( \: log \: x \: ) \: + \: C} \\ \\ \\ (xvi.) \: \: I \: = \int{} \: \dfrac{ \cos(log \: x) \: dx} {x} \\ \\ Let \: \: l og \: x \: \: = \: \: t \\ \: \: \dfrac{1}{x} \: dx \: = \: dt \\ \\ \: Substituting \: in \: I \: , \\ \\ \int{} cos \: t \: dt \: = sin \: t \: + \: C \\ \\ \large {\boxed { I \: \: = \: \: sin \: ( \: log \: x \: ) \: + \: C }}$