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Please solve $$\int\dfrac{1}{\sqrt{\sin x\cos^7x}}dx.$$ Thanks ! In advance for your help . I have already use the $t=\tan x$ in the formula, but it doesn't work well.

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    $\begingroup$ Hi! Since you're a new contributor, I have not downvoted your question for not meeting MSE's community standards. I request you to post more of what you have tried in the future, and where you got stuck, so we can help you better. Indeed, the substitution $u = \tan x$ works and I have posted a solution using it. If you have any questions, please let me know. Don't forget to upvote/accept the answer if you find it helpful. Good day! $\endgroup$ May 15, 2021 at 15:39

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$$\int\dfrac{1}{\sqrt{\sin x\cos^7x}}dx$$ $$={\displaystyle\int}\dfrac{1}{\cos^\frac{7}{2}\left(x\right)\sqrt{\sin\left(x\right)}}\,\mathrm{d}x$$ $$={\displaystyle\int}\class{steps-node}{\cssId{steps-node-1}{\sec^2\left(x\right)}}\cdot\class{steps-node}{\cssId{steps-node-2}{\dfrac{\tan^2\left(x\right)+1}{\sqrt{\tan\left(x\right)}}}}\,\mathrm{d}x$$ Substitute $u = \tan x$, and $du = \sec^2x\ \mathrm dx$ $$={\displaystyle\int}\dfrac{u^2+1}{\sqrt{u}}\,\mathrm{d}x$$ $$={\displaystyle\int}u^\frac{3}{2}\,\mathrm{d}u+{\displaystyle\int}\dfrac{1}{\sqrt{u}}\,\mathrm{d}u$$ $$=\dfrac{2u^\frac{5}{2}}{5}+2\sqrt{u} + C$$ Substituting $\tan x$ back in place of $u$, we get $$=\dfrac{2\tan^\frac{5}{2}\left(x\right)}{5}+2\sqrt{\tan\left(x\right)} + C$$


$$\int\dfrac{1}{\sqrt{\sin x\cos^7x}}dx=\dfrac{2\sqrt{\tan\left(x\right)}\left(\tan^2\left(x\right)+5\right)}{5}+C$$

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