Question

integration of cot dx/3+4 log sinx

Answer 1

Answer:

$\bold{\dfrac{1}{4} \: \: log(3 + 4 \: log \: sinx) \: + \: c}$

Step-by-step explanation:

$\underline{\bold{To \: find}} = > \\ \displaystyle\int \dfrac{cotx}{3 + 4log \: sinx} \: dx$

$\underline{\bold{Concept \: used}} = > \\ 1) \: \dfrac{d}{dx} (sinx) = cosx \\ 2) \: \displaystyle\int\dfrac{dx}{x} = logx$

$\underline{\bold{Solution}} = > \\ \displaystyle\int\frac{cotx}{3 + 4 \: log \: sinx} dx$

$let \\ 3 + 4 \: sin \: logx \: = t \\ differentiating \: both \: sides \\ ( \: 0 + 4 \: \dfrac{cosx}{sinx} \: ) \: dx \: = dt \\ 4 \: cotx \: dx \: = \: dt \\ cotx \: dx \: = \dfrac{dt}{4}$

$I \: = \dfrac{1}{4} \displaystyle\int \dfrac{dt}{t}$

$= \dfrac{1}{4} \: logt \: + \: c$

Answer 2

Answer :

Explanation :

Given,

$\displaystyle{ \sf \: l = \int \: \dfrac{cot \: x}{3 + 4 log(sin \: x) }.dx }$

By Substitution Method,

Let u = 3 + 4log(sin x)

$\sf \:u = 3 + 4 log( sin \: x)$

Differentiating both sides,

$\leadsto \: \sf \: du = \dfrac{4}{sin \: x} dx \: \: \times \dfrac{d(sin \: x)}{dx} \\ \\ \leadsto \: \sf \: du = \dfrac{4cos \: x}{sin \: x} dx \\ \\ \leadsto \: \sf \: du = 4cot \: x.dx \\ \\ \leadsto \: \sf \: dx = \dfrac{du}{4cot \: x}$

The integral can be modified as :

$\longrightarrow \: \displaystyle{ \sf \: l = \int \dfrac{ \cancel{cot \: x}}{u} . \dfrac{du}{ 4 \: \cancel{cot \: x}} } \\ \\ \longrightarrow \: \displaystyle{ \sf \: l = \dfrac{1}{4} \int \: \dfrac{1}{u} .du} \\ \\ \longrightarrow \: \sf \: l = \dfrac{log(u)}{4} + c \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \: l = \frac{ log(3 + 4 log(sin \: x) ) }{4} + c}}$