I am trying to find the integral of $$\int \tan^4 x \, \sec^6 x dx$$

I tried to rewrite as trig identities using $\sec^2 - \tan^2 = 1$ but that got me nowhere so I wrote it like this.

$$\int \frac{\sin^4x}{\cos^4 x} \frac{1}{\cos^6 x} dx$$

$$\int \frac{\sin^4x}{\cos^{10} x} dx$$

Then I use the idea that making u substitutions for cos will get rid of a power of sin so I just say that the power will go $4 3 2 1 0$ and I will get a $- + - + -$ sign change. I am not sure if this is correct or really how this works exactly but I did a few steps of it and it seemed to work correctly.

$$-1\int \frac{1}{u^{10}} dx$$

$$-1\int u^{-10} dx$$

$$-1 \times \frac{u^{-9}}{-9}$$

$$ \frac{u^{-9}}{9}$$

$$ \frac{\cos^{-9}}{9}$$

This is wrong and I am not sure why.

  • $\begingroup$ Did you see how your previous questions were edited? Do you mind using the same format? $\endgroup$ – Gigili Jun 3 '12 at 18:02
  • $\begingroup$ @Gigili What is the difference? $\endgroup$ – toby yeats Jun 3 '12 at 18:08

Your substitution is not correct. You should still still have a $\sin^3 x$ upstairs.

However when integrating a product of an even power of $\tan$ with an even power of $\sec$, you can do the following, which takes advantage of the facts that $\tan^2x+1=\sec^2 x$ and that the derivative of $\tan x$ is $\sec^2 x$:

First write $$\eqalign{\tan^4 x\,\sec^6 x&=\tan^4x\,\sec^4x\cdot \sec^2 x\cr &= \tan^4x \,(\sec^2x)^2\cdot \sec^2 x\cr &=\tan^4x \,(\tan^2 x+1)^2\cdot \sec^2 x.}$$ To evaluate the integral, set $u=\tan x$ (so that $du=\sec^2 x \,dx$).


Here is a relatively easier way. Remember that $\sec^2(x) = 1 + \tan^2(x)$. Hence the integral $$I = \int \tan^4(x) \sec^6(x) dx = \int \tan^4(x) \sec^4(x) \sec^2(x) dx\\ = \int \tan^4(x) \left(1 + \tan^2(x) \right)^2 \sec^2(x) dx $$ Let $\tan(x) = t$. Hence, $\sec^2(x) dx = dt$. Hence, we get that $$I = \int t^4 \left( 1+t^2 \right)^2 dt = \int t^4 \left( 1+2t^2 +t^4 \right) dt = \int \left(t^4 + 2t^6 + t^8 \right) dt\\ =\dfrac{t^5}{5} + \dfrac{2t^7}{7} + \dfrac{t^9}{9} + C$$ Hence, we get that $$I = \dfrac{\tan^5(x)}{5} + \dfrac{2\tan^7(x)}{7} + \dfrac{\tan^9(x)}{9} + C$$


No. You haven't done it correctly. Hint

  • $\tan^{4}(x) \cdot \sec^{4}(x) \cdot \sec^{2}(x)$.

  • $\sec^{4}(x) = (1+\tan^{2}{x})^{2}$.


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