Question

Integral of dx/[4sin²x-sinx] ?

Answer 1

`I=-4/sqrt15ln|(tan(x/2)-4-sqrt15)/(tan(x/2)-4+sqrt15)|-ln|cscx-cotx|+C`

Explanation:

Here,

`I=int1/(4sin^2x-sinx)dx`

`=int1/(sinx(4sinx-1))dx`

`=int((4sinx)-(4sinx-1))/(sinx(4sinx-1))dx`

`=int[4/(4sinx-1)-1/sinx]dx`

`=4int1/(4sinx-1)dx-color(blue)(intcscxdx...tocolor(blue)(Apply(2)`

`=4I_1-color(blue)(ln|cscx-cotx|)...to(A)`

Now ,`I_1=int1/(4sinx-1)dx`

Let, `color(violet)(tan(x/2)=t)=>sec^2(x/2)1/2dx=dt`

`=>(1+tan^2(x/2))dx=2dt=>dx=(2dt)/(1+t^2`

`I_1=int1/(4(2t)/(1+t^2)-1)xx(2dt)/((1+t^2)`

`=int2/(8t-1-t^2)dt`

`=-2int1/(t^2-8t+1)dt`

`=-2int1/((t^2-8t+16-15)dt`

`=-2int1/((t-4)^2-(sqrt15)^2)dt...tocolor(red)(Apply(1)`

`=-2xxcolor(red)(1/(2sqrt15)ln|(t-4-sqrt15)/(t-4+sqrt15)|+c`

Subst. back , `color(violet)(t=tan(x/2)`

`I_1=-1/sqrt15ln|(tan(x/2)-4-sqrt15)/(tan(x/2)-4+sqrt15)|+c`

From `(A) ,` we get

#I=-4/sqrt15ln|(tan(x/2)-4-sqrt15)/(tan(x/2)-4+sqrt15)|-ln|cscx- cotx|+C#

NOTE : Formulas:

`color(red)((1)int1/(x^2-a^2)dx=1/(2a)ln|(x-a)/(x+a)|+c`

`color(blue)((2)intcscxdx=ln|cscx-cotx|+c`