Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
add comment

3 Answers 3

up vote 3 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|improve this answer
    
Thanks,I got it! –  Ave Maleficum Jan 25 '13 at 11:45
add comment

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|improve this answer
1  
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
add comment

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|improve this answer
add comment

Your Answer

 
discard

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.