Question

solve this problem .
I will give you 10 points
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{cos(log {x}^{2}) }{ {x}^{4} } dx$

can be rewritten as

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{cos(2 \: log {x}) }{ {x}^{4} } dx$

$\red{\bigg \{ \because \: log( {x}^{y} ) = y log(x) \bigg \}}$

Let assume that

$\rm :\longmapsto\:I \: = \: \displaystyle\int\tt \dfrac{cos(2 \: log {x}) }{ {x}^{4} } dx$

To evaluate this integral, we use method of Substitution.

$\rm :\longmapsto\:Put \: logx = y$

$\rm :\implies\:x = {e}^{y}$

$\rm :\implies\:dx = {e}^{y} \: dy$

On substituting all these values, we get

$\rm :\longmapsto\:I = \displaystyle\int\tt \dfrac{cos2y}{{e}^{4y}} {e}^{y}dy$

$\rm :\longmapsto\:I = \displaystyle\int\tt \dfrac{cos2y}{{e}^{3y}}dy$

$\rm :\longmapsto\:I = \displaystyle\int\tt {e}^{ - 3y}cos2y \: dy - - - (1)$

now integrating using By parts, we get

$\rm \: = \:{e}^{ - 3y}\displaystyle\int\tt cos2ydy - \displaystyle\int\tt \bigg(\dfrac{d}{dy}{e}^{ - 3y}\displaystyle\int\tt cos2ydy\bigg)dy$

$\rm \: = \: {e}^{ - 3y}\dfrac{sin2y}{2} - \displaystyle\int\tt {e}^{ - 3y}( - 3)\dfrac{sin2y}{2}dy$

$\rm \: = \: \dfrac{{e}^{ - 3y}sin2y}{2} + \dfrac{3}{2}\displaystyle\int\tt {e}^{ - 3y}sin2ydy$

$\rm :\longmapsto\: I = \: \dfrac{{e}^{ - 3y}sin2y}{2} + \dfrac{3}{2}I_1 - - - (2)$

where,

$\rm :\longmapsto\:I_1 = \displaystyle\int\tt {e}^{ - 3y}sin2ydy$

On integrating by parts, we get

$\rm \: = \:{e}^{ - 3y}\displaystyle\int\tt sin2ydy - \displaystyle\int\tt \bigg(\dfrac{d}{dy}{e}^{ - 3y}\displaystyle\int\tt sin2ydy\bigg)dy$

$\rm \: = - \: {e}^{ - 3y}\dfrac{cos2y}{2} + \displaystyle\int\tt {e}^{ - 3y}( - 3)\dfrac{cos2y}{2}dy$

$\rm \: = \: - \dfrac{{e}^{ - 3y}cos2y}{2} - \dfrac{3}{2}\displaystyle\int\tt {e}^{ - 3y}cos2ydy$

$\rm \: = \: - \dfrac{{e}^{ - 3y}cos2y}{2} - \dfrac{3}{2}I$

$\red{\bigg \{ \because \: using \: (1)\bigg \}}$

$\bf\implies \:I_1 = - \: \dfrac{{e}^{ - 3y}cos2y}{2} - \dfrac{3}{2}I - - - (3)$

On substituting equation (3) in equation (2), we get

$\rm :\longmapsto\: I = \: \dfrac{{e}^{ - 3y}sin2y}{2} + \dfrac{3}{2}\bigg( - \dfrac{{e}^{ - 3y}cos2y}{2} - \dfrac{3}{2}I\bigg)$

$\rm :\longmapsto\: I = \: \dfrac{{e}^{ - 3y}sin2y}{2} +\bigg( - \dfrac{3{e}^{ - 3y}cos2y}{4} - \dfrac{9}{4}I\bigg)$

$\rm :\longmapsto\: I = \: \dfrac{{e}^{ - 3y}sin2y}{2} - \dfrac{3{e}^{ - 3y}cos2y}{4} - \dfrac{9}{4}I$

$\rm :\longmapsto\: I + \dfrac{9}{4}I= \: \dfrac{{e}^{ - 3y}sin2y}{2} - \dfrac{3{e}^{ - 3y}cos2y}{4}$

$\rm :\longmapsto\: \dfrac{4 + 9}{4}I= \: \dfrac{{e}^{ - 3y}sin2y}{2} - \dfrac{3{e}^{ - 3y}cos2y}{4}$

$\rm :\longmapsto\: \dfrac{13}{4}I= \: \dfrac{{e}^{ - 3y}sin2y}{2} - \dfrac{3{e}^{ - 3y}cos2y}{4}$

$\rm :\longmapsto\: \dfrac{13}{4}I= \: \dfrac{2{e}^{ - 3y}sin2y - 3{e}^{ - 3y}cos2y}{4}$

$\rm :\longmapsto\: 13 \: I= \: 2{e}^{ - 3y}sin2y - 3{e}^{ - 3y}cos2y$

$\rm :\longmapsto\: 13 \: I= \: {e}^{ - 3y}(2sin2y - 3cos2y)$

$\rm :\longmapsto\: \: I= \dfrac{{e}^{ - 3y}}{13} \:(2sin2y - 3cos2y) + c$

Thus,

$\rm :\longmapsto\: \: I= \dfrac{{x}^{ - 3}}{13} \:(2sin2(logx) - 3cos2(logx)) + c$

Basic Concept used :-

Integration by Parts

✏️See the rule:

• ∫u v dx = u∫v dx −∫u' (∫v dx) dx

• u is the function u(x)

• v is the function v(x)

• u' is the derivative of the function u(x)

For integration by parts , the ILATE rule is used to choose u and v.

where,

• I - Inverse trigonometric functions

• L -Logarithmic functions

• A - Arithmetic and Algebraic functions

• T - Trigonometric functions

• E- Exponential functions

The alphabet which comes first is choosen as u and other as v.