$$ \int \frac{1-33\sec(x)}{\cos(x)-33}dx $$ How do I simplify the fraction using trig identities?
2026-03-25 01:32:03.1774402323
Find the indefinite integral.
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\begin{align*} \int\dfrac{1-33\sec x}{\cos x-33}dx&=\int\dfrac{\sec x(\cos x-33)}{\cos x-33}dx\\ &=\int\sec xdx\\ &=\int\dfrac{\sec x(\sec x+\tan x)}{\sec x+\tan x}dx\\ &=\int\dfrac{\sec^{2}x+\sec x\tan x}{\sec x+\tan x}dx\\ &=\int\dfrac{1}{u}du,~~~~u=\sec x+\tan x\\ &=\log|u|+C\\ &=\log|\sec x+\tan x|+C, \end{align*} where with $u=\sec x+\tan x$, $du=(\sec^{2}x+\tan^{2}x)dx$.