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Evaluate the following integral:

$$\int \frac{\sqrt{\sin ^4x+\cos ^4x}}{\sin ^3x. \cos x }dx$$ where $x \in \big(0,\frac{\pi}{2} \big)$

Could some give me hint as how to approach this question?

I tried to use the fact that $\sin ^4x+\cos ^4x=1-\frac{sin^22x}{2}$ but it didn't help. How should I proceed?

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  • $\begingroup$ Your substitution is right. Now parse this integral into two integrals. In first make a substitution : $cosx = t$ and in second : $sin2x = 2sinxcosx$ $\endgroup$ – openspace Apr 20 '16 at 8:12
  • $\begingroup$ The calculator says that the integral does not converge. $\endgroup$ – Solumilkyu Apr 20 '16 at 8:55
  • $\begingroup$ Did you try $\sin(x)=u$ $\endgroup$ – Archis Welankar Apr 20 '16 at 9:34
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Dividing by $\cos^2x$ on the top and bottom gives $\displaystyle\int\frac{\sqrt{\tan^4x+1}}{\frac{\sin^3x}{\cos x}}dx=\int\frac{\sqrt{\tan^4x+1}}{\tan^3x}\sec^2x \;dx$.

Now let $u=\tan x$ to get $\displaystyle\int\frac{\sqrt{u^4+1}}{u^3}du,\;$ and then let $t=u^2$ to get $\displaystyle\frac{1}{2}\int\frac{\sqrt{t^2+1}}{t^2}dt$.

Next let $t=\tan\theta\;$ to get

$\displaystyle\frac{1}{2}\int\frac{\sec\theta}{\tan^2\theta}\;\sec^2\theta\;d\theta=\frac{1}{2}\int\frac{\sec\theta}{\tan^2\theta}\;(\tan^2\theta+1)\;d\theta=\frac{1}{2}\int\big(\sec\theta+\cot\theta\csc\theta\big)d\theta$

$\displaystyle=\frac{1}{2}\big[\ln\big|\sec\theta+\tan\theta\big|-\csc\theta\big]+C=\frac{1}{2}\left[\ln\left(\sqrt{1+u^4}+u^2\right)-\frac{\sqrt{1+u^4}}{u^2}\right]+C$

$\displaystyle=\frac{1}{2}\left[\ln\left(\sqrt{1+\tan^4x}+\tan^2x\right)-\frac{\sqrt{1+\tan^4x}}{\tan^2x}\right]+C$

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Using your substitution: $$\int{\frac{dx}{sin^{3}x\cdot cosx}-2\int{\frac{cosx dx}{sinx}}}$$

Now using substitution : $t = sinx$: $$\int{\frac{dx}{sin^{3}x\cdot cosx} = \int{\frac{du}{u^{3}(1-u^{2})}}}$$, you could finish it by parsing your integral by some parts. And of course second integral you should know.

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