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How to integrate $\displaystyle\int\frac{\sin ^{2}\theta }{\cos ^{5}\theta }d\theta$? This is also homework,how to start with it?

I try to change into

$$\int\tan ^{2}\theta \sec ^{3}\theta d\theta$$

and then

$$\int\tan ^{}\theta \sec ^{}\theta \; \tan \theta \sec ^{2}\theta d\theta$$

if I set $u=\sec \theta $ then I will deal with a single $ \tan \theta $ then I stuck

Quanto
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Ave Maleficum
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4 Answers4

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$$ \int \dfrac{\sin^2\theta}{\cos^5\theta}\mathscr{d}\theta=\int \dfrac{1}{\cos^5\theta}\mathscr{d}\theta-\int \dfrac{1}{\cos^3\theta}\mathscr{d}\theta$$

Let $I_n:=\int \cos^n\theta\mathscr{d}\theta$

We have
$\begin{align*} I_n=\int \cos^{n-1}\theta\mathscr{d}\sin\theta&=\cos^{n-1}\theta\sin\theta-\int \sin\theta\mathscr{d}\cos^{n-1}\theta \\&=\cos^{n-1}\theta\sin\theta+(n-1)\int\cos^{n-2}\theta\sin^2\theta\mathscr{d}\theta \\&=\cos^{n-1}\theta\sin\theta+(n-1)(I_{n-2}-I_n) \end{align*}$

Hence, $nI_n=(n-1)I_{n-2}+\cos^{n-1}\theta\sin\theta$.

Also, we have $$I_{-1}=\int \sec\theta\mathscr{d}\theta=\ln |\sec\theta+\tan\theta|+C$$

So,

$$I_{-3}=\dfrac{\ln |\sec\theta+\tan\theta|+\sec^2\theta\sin\theta}{2}+C$$ and $$I_{-5}=\dfrac{3I_{-3}+\sec^4\theta\sin\theta}{4}$$
$$\therefore \int \dfrac{\sin^2\theta}{\cos^5\theta}\mathscr{d}\theta=I_{-5}-I_{-3}=\dfrac{-\ln |\sec\theta+\tan\theta|-\sec^2\theta\sin\theta+2\sec^4\theta\sin\theta}{8}+C$$

whoisit
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Shane
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  • Thank you Sir.Although I am sure this question is way beyond my capacity.Will this method be useful for any $\int_{}^{}{\frac{\sin ^{n}\theta }{\cos ^{m}\theta }}d\theta$ ? – Ave Maleficum Jan 25 '13 at 12:34
  • @AveMaleficum: Let $I(n,m)=\int \dfrac{\sin^n\theta}{\cos^m\theta}\mathscr{d}\theta$, I've compute the recurrence relation in similar way: $I(n,m)=\dfrac{\sin^{n+1} \theta}{(m-1)\cos^{m-1}\theta}-\dfrac{n-m+2}{m-1} I(n,m-2)$. You may find similar recurrence relations(with $I(n-2,m)$ or $I(n-2,m-2)$ eg.) by yourself. Try it! ;-) – Shane Jan 25 '13 at 12:53
  • Thank you,Sir.This really helps a lot. – Ave Maleficum Jan 25 '13 at 13:06
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  1. $\int\sec{x} \ dx=\ln|\sec x+\tan x| , \int \sec^2{x} \ dx=\tan x$ .
  2. $\int\dfrac{\sin^2 x}{\cos^2 x} \ dx=\int\sec^3 x+\sec^5 x \ dx$ .

If $I=\int \sec^5 x \ dx=\int \sec^3 \sec^2 x \ dx=\int \sec^3 (\tan x)' \ dx$
then $I=\sec^3 x\tan x -\int3\sec^2x \sec x\tan x\tan x \ dx =\\ \sec^3 x\tan x -\int3\sec^3x \tan^2 x=\sec^3 x\tan x -3I+3\int\sec^3x \ dx...$

Do the same for $\int \sec^3 x$ and use 1.

P..
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  • The integrals of the odd powers of secant are not all that easy. I think they take more than "integration by parts". – Gerry Myerson Jan 25 '13 at 12:32
  • @GerryMyerson: I believe now is better. Is there a difference for the integrals of even powers of secant? I believe there isn't. For any power you can calculate $\int \sec^n x \ dx$ recursively. – P.. Jan 25 '13 at 12:49
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    Even powers are much easier because they can be written as $P(\tan\theta)d(\tan\theta)$ for some polynomial $P$. – Gerry Myerson Jan 25 '13 at 22:58
  • You are right @GerryMyerson. Thanks. – P.. Jan 26 '13 at 06:41
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for this type of problems, there are some ways of doing it.

you can do this by doing inverse trig identities like setting $\theta$ = $\arcsin(u)$ and don't forget $d\theta$ = $\frac{1}{\sqrt{1-x^2}}$ $du$

then you will get a integral will look like $\int \frac{u^2}{(1-u^2)^3} du$ then you can finish it off but remember your $\theta$ = $\arcsin(u)$ will be now u = $\sin(\theta)$ when you get your answer.

EM4
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It is much simpler to integrate with the substitution $\tan \theta= \sinh t$ $$\int{\frac{\sin ^{2}\theta }{\cos ^{5}\theta }d\theta } =\frac14\int \sinh^22t \ dt=\frac1{32}\sinh 4t-\frac t8 +C $$

Quanto
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