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I can't find a way to solve this: $$\int\frac{\cos^5(x)}{\sin(x)^{1/2}}dx$$ I'm not sure how i have to proceed, can you help me? I've tried the substitution u=(sinx), but it got me nowhere.

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  • $\begingroup$ Have you tried the Weierstrass substitution? $\endgroup$ – Lucian Apr 5 '14 at 17:25
  • $\begingroup$ Is sin(x)^1/2: $\sin(x^{1/2})$ or $\sqrt{\sin(x)}$? $\endgroup$ – user122283 Apr 5 '14 at 17:29
  • $\begingroup$ it is sin(x)−−−−−√ $\endgroup$ – Wollacy Silva Apr 5 '14 at 17:37
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\begin{align*} \int \frac{\cos^5 x}{\sin^{1/2} x} \,dx &= \int \frac{(1-u^2)^2}{\sqrt u} \,du &&\text{($u=\sin x$)} \\ &= \int 2(1-v^4)^2 \,dv &&\text{($v = \sqrt u$)} \end{align*} Now it's just a polynomial; expand the square, integrate term by term, and replace $v$ with the equivalent in $x$.

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Hint: $$ \frac{\cos^5(x)}{\sin(x)^{1/2}} = \frac{\cos(x)(1-\sin(x)^2)^2}{\sin(x)^{1/2}}. $$ Now use the substitution $u=\sin x$.

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Hint:

(Long) METHOD 1: See that $$\dfrac{d\sqrt{\sin x}}{dx}=\dfrac{1}{2\sqrt{\sin x}}\cos x\\ \implies \dfrac{1}{\sqrt{\sin x}}\cos xdx=2d\sqrt{\sin x}$$ Thus, $$\int\dfrac{\cos^5x}{\sqrt{\sin x}}dx=2\int{\cos^4x}d\sqrt{\sin x}=\\ 2\left(\sqrt{\sin x}\cos^4x+4\int\sqrt{\sin x}\cos^3 x\sin xdx\right)$$ Now, $\int\cos^3 x\sin^{3/2} xdx$ can be calculated by $\dfrac{d\sin^{5/2}x}{dx}=\dfrac{5}{2}\sin^{3/2}x\cos x$, i.e., $\sin^{3/2}x\cos xdx=\dfrac{2}{5}d\sin^{5/2}x$. Continue integration by parts.

(Shorter) METHOD 2: Use the identity $1-\sin^2 =\cos^2$, to get $$\int\dfrac{(1-\sin^2x)^2}{\sqrt{\sin x}}dx=2\int2{(1-u^4)^2}du,\text{ $u=\sqrt{\sin x}$}\\ \implies \int\dfrac{(1-\sin^2x)^2}{\sqrt{\sin x}}dx=\int(1+u^8-2u^4)du=\\ \boxed{\sqrt{\sin x}+\dfrac{\sqrt{\sin^9 x}}{9}-2\dfrac{\sqrt{\sin^5x}}{5}+C}$$

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