Consider the integral $$\int \sec^4(x)\tan(x)$$
Now right off the bat I see two ways of solving this.
- Let u=$\sec(x)$
2.Use integration by parts
Now doing the first way results in the integrand looking like$$\int u^3du=\frac{1}{4}\sec^4(x)+C $$
Which is correct but it's not the answer I'm looking for, so instead we'll do it the second way.
$$\int \sec^2(x)\cdot\sec^2(x)\tan(x)dx$$$$\int\left(\tan^2(x)+1\right)\sec^2(x)\tan(x)dx $$$$\int\sec^2(x)\tan^3(x)+\sec^2(x)\tan(x)dx$$Now this is where I got stuck, because I don't know whether to continue with Pythagorean identities or to factor a term out and solve for that. Or perhaps even break the two up and create two integrals.
$\endgroup$ 54 Answers
$\begingroup$Write your integrand in the form$$\tan(x)(\tan^2(x)+1)\sec^2(x)$$ and substitute $$u=\tan(x)$$ and you will get $$\int u(u^2+1)\,du$$
$\endgroup$ 4 $\begingroup$Hint:
Bioche's rules suggest to use the substitution $u=\tan x,\;\mathrm d u=\sec^2 x\,\mathrm dx$ to obtain a polynomial in $u$.
$\endgroup$ 1 $\begingroup$The form of the answer you want says you have$$ \int \tan^3(x) + \tan(x) \,\mathrm{d}(\tan x) \\ = \int \left( \tan^3(x) + \tan(x) \right) \sec^2(x) \,\mathrm{d}x$$ in the prior step, which you do. Let $u = \tan x$.
$\endgroup$ $\begingroup$I think the solution to your question is given in the image above.
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