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 evaluate $$\int \frac{\sin^3 x}{\cos^5x}dx\ ?$$

I've tried various substitutions with $\sin x = u$ or $\cos x = u$, I've tried using Euler's formula which result in too heavy calculations and I've tried using $\sin^2x + \cos^2x = 1$ in various forms without success.

share|improve this question
3  
You tried $u=\cos x$? What was the output? –  Did May 2 at 20:46

3 Answers 3

up vote 13 down vote accepted

Hint: $$\int \frac{\sin^{3}x}{\cos^{5}x}dx=\int \frac{(1-\cos^{2}x)\sin x}{\cos^{5}x}dx$$ and try substitution $$t=\cos x.$$

share|improve this answer

Another approach: $$ \begin{align} \int \frac{\sin^3 x}{\cos^5x}dx&=\int \frac{\sin^3 x}{\cos^3x\cos^2x}dx\\ &=\int\left(\frac{\sin x}{\cos x}\right)^3\frac{dx}{\cos^2x}\\ &=\int\tan^3x\ d(\tan x)\\ &=\frac14\tan^4x+C. \end{align} $$

share|improve this answer

Let $u=\cos x$ then $du=-\sin xdx$ so $$\int\frac{\sin^3x}{\cos^5x}dx=-\int\frac{1-u^2}{u^5}du=\frac{u^{-4}}{4}-\frac{u^{-2}}{2}+C=\frac{\cos^{-4}x}{4}-\frac{\cos^{-2}x}{2}+C$$

share|improve this answer

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.