Question

evaluate the integral. (use c for the constant of integration.) ∫ tan^3(x)sec^6(x)dxI am stuck on the first step.According to my book, it says "if the power of the secant is even, save a factor of sec^2(x) and use sec^2(x) = 1 + tan^2(x) to express the remaining factors in terms of tan(x)If the power of tangent is odd, save a factor of sec(x)tan(x) and use tan^2(x) = sec^2(x) - 1 to express the remaining factors in terms of sec(x).The power of tangent is odd and the power of secant is even, what should I do next? Can I apply both rules to the equation? Or will it be wrong? Please show steps and help guide me.

Answer 1

`\displaystyle {(sec^8(x))/(8)-(sec^6(x))/(6)+C}`

Explanation:

`\displaystyle \int {\tan^3(x)\sec^6(x)} \, dx\\\\=\int {\tan^2(x)\sec^5(x)\tan(x)\sec(x)} \, dx\\=\int {\bigr(\sec^2(x)-1\bigr)\sec^5(x)\tan(x)\sec(x)} \, dx\\`

Now let
`u=\sec(x)` and
`du=\tan(x)\sec(x)dx`:

`\displaystyle \int {\bigr(u^2-1\bigr)u^5} \, du\\\\=\int {u^7-u^5} \, du\\\\=(u^8)/(8)-(u^6)/(6)+C\\\\=\bold{(sec^8(x))/(8)-(sec^6(x))/(6)+C}`

So, because the tangent term was odd and the secant term was even, we had to break it up so we could have the factor sec(x)tan(x), and then turn tan²(x) into sec²(x)-1. Also, because the derivative of sec(x) is sec(x)tan(x), the substitution becomes really simple as you see above.