# Evaluate the following integral integrate of ((tanx)^3)((secx)^6) dx

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### 1 Answer

You need to use the trigonometric identity

`sec^2 x = 1(/cos^2 x) = 1 + tan^2 x` such that:

`int tan^3 x sec^6 x dx = int tan^3 x(1 + tan^2 x)^2 sec^2 x dx`

You should use substitution method such that:

`tan x = t => sec^2 x dx = dt`

You need to substitute the variable of integrand yields:

`int tan^3 x(1 + tan^2 x)^2 sec^2 x dx = int t^3(1+t^2)^2 dt`

Expanding the square in the integrand yields:

`int t^3(1+t^2)^2 dt = int (t^3 + 2t^5 + t^7) dt`

Using the property of linearity of integral such that:

`int (t^3 + 2t^5 + t^7) dt = int t^3 dt + int 2t^5 dt + int t^7 dt`

`int (t^3 + 2t^5 + t^7) dt = t^4/4 + 2t^6/6 + t^8/8 + c`

`int (t^3 + 2t^5 + t^7) dt = t^4/4 + t^6/3 + t^8/8 + c`

Substituting back tan x for t yields:

`int tan^3 x sec^6 x dx = (tan^4 x)/4 + (tan^6 x)/3 +(tan^8 x)/8 + c`

**Hence, evaluating the indefinite integral using trigonometric identites and substitution method yields `int tan^3 x sec^6 x dx = (tan^4 x)/4 + (tan^6 x)/3 +(tan^8 x)/8 + c. ` **