# Integral of $\int{\tan^{5}(x)\sec^4(x)}dx$?

Here's my attempt at the problem: $$\int{\tan^{5}(x)\sec^4(x)}dx= \int{\frac{\sin^5(x)}{\cos^9(x)}}\,dx= \int{\frac{\sin(x)(\sin^2(x))^2}{\cos^9(x)}}\,dx= \int{\frac{\sin(x)(1-\cos^2(x))^2}{\cos^9(x)}}\,dx= \int{\frac{(1-u^2)^2}{u^9}}\,du= \int{\frac{u^4-2u^2+1}{u^9}}\,du= \int{\Big(\frac{1}{u^5}-\frac{2}{u^7}+\frac{1}{u^9}}\Big) \,du= -\frac{1}{4u^4}+\frac{1}{3u^6}-\frac{1}{8u^8}+C= -\frac{\sec^4(x)}{4}+\frac{\sec^6(x)}{3}-\frac{\sec^8(x)}{8}+C$$

It seems, however, that the actual answer should be: $$\frac{\sec^4(x)}{4}-\frac{\sec^6(x)}{3}+\frac{\sec^8(x)}{8}+C$$

What am I doing wrong?

• $$d( \cos x ) = - \sin x dx$$ – ILoveMath Dec 11 '16 at 19:55
• So should $\int{\frac{(1-u^2)^2}{u^9}}du$ instead be $-\int{\frac{(1-u^2)^2}{u^9}}du$ in the above solution? – kylemart Dec 11 '16 at 20:01
• yesssssssssssssssss – ILoveMath Dec 11 '16 at 20:02
• Ok, Remember, this: $$\int_a^b f(x) dx = - \int_b^a f(x) dx$$ – ILoveMath Dec 11 '16 at 20:08
• @LanierFreeman Yeah. My textbook uses the tan / sec relation to solve the problem, but I was curious to see if it could be done easily with sin / cos – kylemart Dec 11 '16 at 20:38

Substituting $u=\tan(x),\,du=\sec^2(x)\,dx$