What is the integral of tan^{-8}(x)sec^4(x) ?

The integral of tan^{-8}(x)sec^4(x) is -1/7 cot^7(x)-1/5 cot^5(x)+C

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integral of tan^{-8}(x)sec^4(x)

\int tan^{- 8} (x) sec^{4} (x) d x

- \frac{1}{7} cot^{7} (x) - \frac{1}{5} cot^{5} (x) + C

Solution steps

\int tan^{- 8} (x) sec^{4} (x) d x

Rewrite using trig identities

= \int (1 + tan^{2} (x)) sec^{2} (x) tan^{- 8} (x) d x

Apply u-substitution

: \int \frac{1 + u ^{2}}{u ^{8}} d u

= \int \frac{1 + u ^{2}}{u ^{8}} d u

Expand

\frac{1 + u ^{2}}{u ^{8}} : \frac{1}{u ^{8}} + \frac{1}{u ^{6}}

= \int \frac{1}{u ^{8}} + \frac{1}{u ^{6}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u ^{8}} d u + \int \frac{1}{u ^{6}} d u

\int \frac{1}{u ^{8}} d u = - \frac{1}{7 u ^{7}}

\int \frac{1}{u ^{6}} d u = - \frac{1}{5 u ^{5}}

= - \frac{1}{7 u ^{7}} - \frac{1}{5 u ^{5}}

Substitute back

u = tan (x)

= - \frac{1}{7 tan ^{7} ( x )} - \frac{1}{5 tan ^{5} ( x )}

Simplify

- \frac{1}{7 tan ^{7} ( x )} - \frac{1}{5 tan ^{5} ( x )} : - \frac{1}{7} cot^{7} (x) - \frac{1}{5} cot^{5} (x)

= - \frac{1}{7} cot^{7} (x) - \frac{1}{5} cot^{5} (x)

Add a constant to the solution

= - \frac{1}{7} cot^{7} (x) - \frac{1}{5} cot^{5} (x) + C

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What is the integral of tan^{-8}(x)sec^4(x) ?
The integral of tan^{-8}(x)sec^4(x) is -1/7 cot^7(x)-1/5 cot^5(x)+C