Question

sin x/sin 4x ,Integrate the given function defined on proper domain w.r.t. x.

Answer 1

HELLO DEAR,




GIVEN:-
∫sinx/sin4x.dx

=> I = ∫sinx/(2sin2xcos2x).dx

=> I = ∫sinx/(4sinxcosxcos2x).dx

=> I = ∫dx/(4cosxcos2x)

=> I = ∫dx/2(2cosxcos2x)

=> I = ∫dx/2(cos3x + cosx)

=> I = 1/2∫sec3x.dx + 1/2∫secx

=> I = 1/6log|sec3x + tan3x| + 1/2log|secx + tanx| + C.


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

we have to integrate $\int{\frac{sinx}{sin4x}}\,dx$
we know, sin2A = 2sinA.cosA
so,$\int{\frac{sinx}{2sin2x.cos2x}}\,dx$

= $\int{\frac{sinx}{4sinx.cosx.cos2x}}\,dx$

=$\int{\frac{1}{4cosx.cos2x}}\,dx$

=$\frac{1}{4}\int{\frac{secx}{cos2x}}\,dx$

we know, cos2x = (1 - tan²x)/(1 + tan²x) use it here,

= $\frac{1}{4}\int{\frac{secx(1+tan^2x)}{1-tan^2x}}\,dx$

=$\frac{1}{4}\int{\frac{secx.sec^2x}{(1-tan^2x)}}\,dx$

Let tanx = z
differentiate both sides,
sec²x.dx = dz

so, $\frac{1}{4}\int{\frac{\sqrt{z^2-1}}{1-z^2}}\,dz$

= $\frac{-1}{4}\int{\sqrt{z^2-1}}\,dz$

= $\frac{-1}{8}[z\sqrt{z^2-1}-log|\sqrt{z^2-1}+z|]+C$

now put z = tanx

= $\frac{1}{8}[tanx\sqrt{tan^2x-1}-log|\sqrt{tan^2x-1}+tanx]+C$