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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?

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    $\begingroup$ $$d( \cos x ) = - \sin x dx $$ $\endgroup$ – ILoveMath Dec 11 '16 at 19:55
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    $\begingroup$ 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? $\endgroup$ – kylemart Dec 11 '16 at 20:01
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    $\begingroup$ yesssssssssssssssss $\endgroup$ – ILoveMath Dec 11 '16 at 20:02
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    $\begingroup$ Ok, Remember, this: $$ \int_a^b f(x) dx = - \int_b^a f(x) dx $$ $\endgroup$ – ILoveMath Dec 11 '16 at 20:08
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    $\begingroup$ @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 $\endgroup$ – kylemart Dec 11 '16 at 20:38
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Substituting $u=\tan(x),\,du=\sec^2(x)\,dx$

\begin{eqnarray} \int\tan^5(x)\sec^4(x)\,dx&=&\int\tan^5(x)\sec^2(x)\sec^2(x)\,dx\\ &=&\int\tan^5(x)\left[\tan^2(x)+1\right]\sec^2(x)\,dx\\ &=&\int u^5\left(u^2+1\right)\,du\\ &=&\int u^7+u^5\,du\\ &=&\frac{1}{8}u^8+\frac{1}{6}u^6+c\\ &=&\frac{1}{8}\tan^8(x)+\frac{1}{6}\tan^6(x)+c \end{eqnarray}

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