# How to integrate $\int_{}^{}{\frac{\sin ^{3}\theta }{\cos ^{6}\theta }d\theta }$?

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.

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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.

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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

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

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

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

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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.

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