Solving Integrals with Even Powers of Trigonometric Functions

[alba_ei]

Hello I have this integral and I can't solve it


\int\csc^6 x dx


Well i first start with trig identnity \csc x = \frac{1}{sen x} so my first attempt to slove it looks like this

\int\frac{dx}{(\sin^2 x)^3}

then i subsituted with \sin^2 x = \frac{1 - \cos 2x}{2}

and i get

8\int\frac{dx}{(1-cos 2x)^3} and i saw that i was going to

nowhere so i try another way

i took the function and use the trig identity to get this

\int\csc^6 x dx =\int \frac{\sec^6 x}{\tan^6 x} dx

u = \tan x du = \sec^2 x

then

\int\frac{du^3}{tan^6 x}

and i get stucked again. any ideas?

 

[neutrino]

i took the function and use the trig identity to get this

\int\csc^6 x dx =\int \frac{\sec^6 x}{\tan^6 x} dx

That looks most promising.

If u = \tan{x}, then du = \sec^2{x}dx (Don't forget the dx)

Now, rewrite the integral as

\int{\frac{\sec^4{x}\sec^2{x}dx}{\tan^6{x}}}

Now, think of an identity that relates \sec{x} and \tan{x}.

 

[dextercioby]

Make \tan x =t and then u'll get 3 simple integrals.

 

[VietDao29]

Hello I have this integral and I can't solve it\int\csc^6 x dx

Here's the trick, if the power of csc function is even, you can use the u-substitution u = cot(x).
And if the power of sec function is even, you can use the u-substitution u = tan(x)
So, in your problem, the power of csc function is even, we use the u-substitution: u = cot(x)
du = -csc²(x) dx, so your integral will become:
\int \csc ^ 6 (x) dx = \int ( \csc ^ 4 (x) \times \csc ^ 2 (x) ) dx = \int {( \csc ^ 2 (x) )} ^ 2 \times \csc ^ 2 (x) dx = \int {\left( 1 + \cot ^ 2 (x) \right)} ^ 2 \times \csc ^ 2 (x) dx
= -\int {(1 + u ^ 2)} ^ 2 du = ...
Can you go from here? :)