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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.

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

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

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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 $x$) $$\frac{d}{dx}(f(x)^n)=nf(x)^{n-1}f'(x)$$

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

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