Question

1-cos3x/sin²3x,Integrate the given function defined on proper domain w.r.t. x.

Answer 1

Dear Student,

Solution:

Answer of your problem is attached .

Steps of Integration:

1) first apply linearity

2) Now substitute 3x = t

on differentiation ,we get 3dx = dt

So, dx = dt/3

3) Apply substitution ,and linearity

4) After applying linearity ,in first term you found cos t/ sin^2 t ; assume sin t = u

so,on differentiating we get cost dt = du

5) Apply substitution, you will get that simple form ,which you can easily integrate.

6) Second term converted into cosec^ 2 t, which is integration of - cot t

7) Redo substitution, you will get the final answer.

Hope it helps you

Answer 2

question is $\int{\frac{1-cos3x}{sin^23x}}\,dx$

we know, $1-cos2A=2sin^2A$
$sinA=2sin\frac{A}{2}.cos\frac{A}{2}$
so, $1-cos3x=2sin^2\frac{3x}{2}$
$sin^23x=\left(sin(3x/2).cos(3x/2)\right)^2$
=$4sin^2(3x/2).cos^2(3x/2)$

now,
$\int{\frac{2sin^2\frac{3x}{2}}{4sin^2(3x/2).cos^2(3x/2)}}\,dx$

$=\int{\frac{1}{2}\frac{1}{cos^2\frac{3x}{2}}}\,dx\\\\\\=\frac{1}{2}\int{sec^2\frac{3x}{2}}\,dx\\\\\\=\frac{1}{2}\left[\frac{tan\frac{3x}{2}}{\frac{3}{2}}\right]+C\\\\\\=\frac{1}{3}tan\frac{3x}{2}+C$