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Question Number 119970 by bramlexs22 last updated on 28/Oct/20

$$\int \frac{dx}{x^2\sqrt{25-x^2}}?$$

Answered by bemath last updated on 28/Oct/20

$$\int \frac{dx}{x^3\sqrt{\frac{25}{x^2}-1}}dx=\int \frac{x^{-3}}{\sqrt{25x^{-2}-1}}dx$$$$lettingm=25x^{-2}-1\Rightarrow dm=-50x^{-3}dx$$$$\int \frac{-\frac{dm}{50}}{\sqrt{m}}=-\frac{1}{50}\int m^{-\frac{1}{2}}dm$$$$=-\frac{1}{25}\sqrt{m}+c=-\frac{1}{25}\sqrt{25x^{-2}-1}+c$$$$=-\frac{\sqrt{25-x^2}}{25x}+c$$

Answered by Dwaipayan Shikari last updated on 28/Oct/20

$$\int \frac{dx}{25{sin}^2\theta \sqrt{25-25{sin}^2\theta }}x=5sin\theta \Rightarrow 1=5cos\theta \frac{d\theta }{dx}$$$$=\int \frac{5cos\theta d\theta }{25{sin}^2\theta .5cos\theta }$$$$=\frac{1}{25}\int \frac{1}{{sin}^2\theta }$$$$=-\frac{1}{25}{\left(tan\theta \right)}^{-1}=-\frac{1}{25}{(.\frac{sin\theta }{\sqrt{1-{sin}^2\theta }})}^{-1}=-\frac{1}{25}{(.\frac{5x}{5\sqrt{25-x^2}})}^{-1}=-\frac{1}{25}.{\left(\frac{x}{\sqrt{25-x^2}}\right)}^{-1}+C$$$$=-\frac{1}{25}\left(\frac{\sqrt{25-x^2}}{x}\right)+C$$

Commented by bemath last updated on 28/Oct/20

$$\cot \theta =\frac{\cos \theta }{\sqrt{1-\cos {}^2\theta }}$$

Commented by Dwaipayan Shikari last updated on 28/Oct/20

$$Isupposeitastan\theta typo$$$$Thankingforcorrection$$

Answered by Bird last updated on 28/Oct/20

$$I=\int x^{-2}{(25-x^2)}^{-\frac{1}{2}}dxbyparts$$$$I=-\frac{1}{x}{(25-x^2)}^{-\frac{1}{2}}-\int -\frac{1}{x}(-\frac{1}{2})(-2x){(25-x^2)}^{-\frac{3}{2}}$$$$=-\frac{1}{x\sqrt{25-x^2}}+\int \frac{dx}{(25-x^2)\sqrt{25-x^2}}$$$$\int \frac{dx}{(25-x^2)\sqrt{25-x^2}}{=}_{x=5sint}$$$$=\int \frac{5costdt}{25{cos}^{2}t.5.cost}$$$$=\frac{1}{25}\int \frac{dt}{\frac{1+cos\left(2t\right)}{2}}=\frac{2}{25}\int \frac{dt}{1+cos\left(2t\right)}$$$${=}_{tant=u}\frac{2}{25}\int \frac{du}{(1+u^2)(1+\frac{1-u^2}{1+u^2})}$$$$=\frac{2}{25}\int \frac{du}{1+u^2+1-u^2}=\frac{1}{25}\int du$$$$=\frac{u}{25}+c=tan\left(arcsin\left(\frac{x}{5}\right)\right)+c\Rightarrow$$$$I=-\frac{1}{x\sqrt{25-x^2}}+tan(arcsin\left(\frac{x}{5}\right)+C$$

Commented by Bird last updated on 28/Oct/20

$$I=-\frac{1}{x\sqrt{25-x^2}}+\frac{1}{25}arctan\left(arcsin\left(\frac{x}{5}\right)\right)+C$$

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