Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to integrate $\int_{}^{}{\frac{\sin ^{3}\theta }{\cos ^{6}\theta }d\theta }$? This is kind of homework,and I have no idea where to start.

share|cite|improve this question
up vote 4 down vote accepted

You can rewrite it as $$\int\tan^3\theta \sec^3\theta d\theta.$$ Note that $d(\sec\theta)=\tan\theta\sec\theta d\theta$ and $\tan^2\theta=\sec^2\theta-1$, we have $$\int\tan^3\theta \sec^3\theta d\theta=\int \tan^2\theta \sec^2\theta d(\sec\theta) =\int (\sec^2\theta-1) \sec^2\theta d(\sec\theta).$$ Now if we let $t=\sec\theta$, then....I leave the remaining parts to you.

share|cite|improve this answer
Thanks,I got it! – Ave Maleficum Jan 25 '13 at 11:45

Or We can also Calculate it Using Trig. Substitution

$\displaystyle \int\frac{\sin^3 \theta}{\cos^6 \theta}d\theta = \int\frac{1-\cos^2 \theta}{\cos^6 \theta} \times \sin \theta d\theta$

Let $\cos \theta = t $ and $\sin \theta d\theta = -dt$ So Integral is

$\displaystyle \int\frac{t^2-1}{t^6}dt = \int t^{-4}dt-\int t^{-6}dt$

$\displaystyle = -\frac{1}{3}t^{-3}+\frac{1}{5}t^{-5}+\mathbb{C}$

$\displaystyle = -\frac{1}{3}(\cos t )^{-3}+\frac{1}{5}(\cos t) ^{-5}+\mathbb{C}$

share|cite|improve this answer
Is there a reason you have used $\mathbb C$ to denote the constant? It seems rather peculiar to me. – user50407 Jan 25 '13 at 22:23

One way is to avoid cumbersome calculations by using s for sine and c for cosine. Split the s^3 in the numerator into s*s^2, using s^2 = 1 - c^2 and putting everything in place you have

s*(1-c^2)/c^6. Since the lonley s will serve as the negative differential of c, the integrand reduces nicely into (1-c^2)*(-dc)/c^6. Divide c^6 into the numerator and you have c^(-6)-c^(-4) which integrates nicely using the elementary power rule, the very first integration rule you learned! Of course, don't forget to backtrack, replacing the c's with cos() and you are done. No messy secants and tangents and certainly no trig substitutions.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.