Evaluate the integral:

$\displaystyle \int_{\pi/6}^{\pi/2} \frac{\cos(x)}{\sin^{5/7}(x)}\, dx$

(using substitution)

Here's my attempt at solution:

u = $\sin^5(x)$

$du = 5\sin^4(x) \cdot \cos(x) \cdot dx$

$ \frac {1}{5\sin^4(x)} du = \cos(x) \cdot dx $

Also, lower and upper limits for integration will be different for new variable u

since $u = \sin^5(x)$

new lower limit is $\frac {1}{32}$

and upper limit is $1$

Making substitution:

$\displaystyle \int_{\frac{1}{32}}^{1} \frac{1}{u^\frac{1}{7}} \cdot \frac{1}{5\sin^4(x)} du$

... and I'm stuck, I dunno how to take an integral of $\frac{1}{5\sin^4(x)} $

Solution shouldn't come to this, there must be another way.

  • $\begingroup$ Let $u=\sin x$. The motivation is to recall what you do when you apply chain rule on differentiation. $\endgroup$ – MonkeyKing Apr 27 '15 at 6:35
  • $\begingroup$ @MonkeyKing I thought I was supposed to define as variable everything that's under a root. $\endgroup$ – dramadeur Apr 27 '15 at 6:36
  • $\begingroup$ Unfortunately, there's not just one way you have to make a substitution. Every integral is different. The more you do, though, the better/faster you'll be able to find the easiest sub to make. $\endgroup$ – Curious Apr 27 '15 at 6:42

You may just perform the change of variable $$v=\sin x, \qquad dv=\cos x\:dx,$$ giving

$$ \int_{\pi/6}^{\pi/2} \frac{\cos(x)}{\sin^{5/7}(x)}\, dx=\int_{1/2}^{1} \frac{dv}{v^{5/7}}=\left[\frac72 v^{2/7}\right]_{1/2}^1=\frac{7}{2}\left(1-\frac{1}{2^{2/7}}\right). $$


Hint: $\cos x$ is the derivative of $\sin x$, and use the chain rule, where (the supersrcipt $n$ is the power, and $f'(x)$ is the derivative of $f(x)$ with respect to $x$) $$\frac{d}{dx}(f(x)^n)=nf(x)^{n-1}f'(x)$$


Try an easier $u$ sub. Hint: Try the one you used, but exclude any exponents.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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