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Question Number 121203 by benjo_mathlover last updated on 05/Nov/20

$$\int \frac{\cos {}^5\left(\mathrm{x}\right)}{\sqrt{\sin \left(\mathrm{x}\right)}}\mathrm{dx}$$

Answered by MJS_new last updated on 06/Nov/20

$$\int \frac{{\cos }^{5}x}{\sqrt{\sin x}}dx=$$$$[t=\sqrt{\sin x}\to dx=\frac{2\sqrt{\sin x}}{\cos x}dt]$$$$=2\int {(t^4-1)}^{2}dt=$$$$=\frac{2}{9}{t}^9-\frac{4}{5}{t}^5+2t=$$$$=2(\frac{1}{9}{\sin }^{4}x-\frac{2}{5}{\sin }^{2}x+1)\sqrt{\sin x}+C$$

Answered by bobhans last updated on 06/Nov/20

$$\int \frac{\cos {}^4\left(x\right)\cos \left(x\right)}{\sqrt{\sin \left(x\right)}}dx=\int \frac{{(1-\sin {}^2\left(x\right))}^{2}d\left(\sin x\right)}{\sqrt{\sin \left(x\right)}}$$$$let\sqrt{\sin x}=u\Rightarrow \sin x=u^2\wedge d\left(\sin x\right)=2udu$$$$\int \frac{{(1-u^4)}^2\left(2udu\right)}{u}=2\int (u^8-2u^4+1)du$$$$=2(\frac{u^9}{9}-\frac{2u^5}{5}+u)+c$$$$=2(\frac{\sqrt{\sin {}^9\left(x\right)}}{9}-\frac{2\sqrt{\sin {}^5\left(x\right)}}{5}+\sqrt{\sin \left(x\right)})+c$$

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