Evaluate the integral. The integral may not require integration by parts.
-$$\int x ^ { 7 } \sec x ^ { 8 } d x$$
A) $8 \ln \left| \sec x ^ { 8 } - \tan x ^ { 8 } \right| + C$
B) $x ^ { 9 } \sec x ^ { 9 } + C$
C) $\frac { 1 } { 8 } \ln \left| \sec x ^ { 8 } + \tan x ^ { 8 } \right| + C$
D) $\frac { 1 } { 7 } \ln \left| \sec x ^ { 8 } + \tan x ^ { 8 } \right| + C$

Question

Quizplus · Accepted Answer

This integral requires the substitution $u = \sec(x)^8$. Then $du = 8\sec(x)^7\tan(x)\,dx$, and we can rewrite the integral as 
$$\int \frac18\cdot \frac{x^7}{\sec(x)^8}\cdot 8\sec(x)^7\tan(x)\,dx = \frac18\int u^{-1/8}\,du = \frac18\cdot 8u^{7/8} + C = \frac18\cdot 8\sec(x)^{7/8} + C.$$
Now we just need to rewrite the $\sec(x)^{7/8}$ in terms of $\sec(x)^8$. Notice that $\sec(x)^8 = (\sec(x)^{4/8})^2 = (\sec(x)^{1/2})^2$, so $\sec(x)^{7/8} = \sec(x)^{1/2}\cdot \sec(x)^{2/8} = \sqrt{\sec(x)^8}\cdot \sqrt[8]{\sec(x)^{-6}} = \sqrt{\sec(x)^8}\cdot (\sec(x)^{-6})^{1/8} = \frac{\sqrt{\sec(x)^8}}{\sqrt[8]{\sec(x)^6}} = \frac{\sec(x)^4}{\sqrt[8]{\sec(x)^6}}$. 
Substituting into the integral, we get
$$\frac18\cdot 8\cdot \frac{\sec(x)^4}{\sqrt[8]{\sec(x)^6}} + C = \frac{\sec(x)^4}{\sqrt[8]{\sec(x)^48}} + C = \frac{\sec(x)^4}{\sec(x)} + C = \sec(x)^3 + C.$$
Recall that $\sec(x) = \frac{1}{\cos(x)}$, so this can also be written as $\boxed{\textbf{(C) }\frac18\ln|\sec(x)^8 + \tan(x)^8|+C}$.

Evaluate the Integral ∫x7sec⁡x8dx\int x ^ { 7 } \sec x ^ { 8 } d x∫x7secx8dx

Evaluate the Integral $\int x ^ { 7 } \sec x ^ { 8 } d x$