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I have tried this integral with $\cos$ substitution, but I don't understand why it's wrong. So could you please tell me which step is wrong? Here are my steps: $$ x=5\cos\theta\\ dx=-5\sinθ\,dθ\\ $$ therefore \begin{align} &(1)=\int\sqrt{25-25\cos^2θ}\left(25\cos^2θ\right)(-5\sinθ)\,dθ\\ &(2)=-5^4\int\sin^2θ\cos^2θ\,dθ\\ &\qquad\left(\sin2θ=2\sinθ\cosθ\implies\sin^22θ=4\sin^2θ\cos^2θ\right)\\ &(3)=\frac{-5^4}4\int\sin^22θ\,dθ\\ &\qquad\left(\sin^2θ=\frac12(1-\cos2θ)\implies\sin^22θ=\frac12(1-\cos4θ)\right)\\ &(4)=\frac{-5^4}8\int(1-\cos4θ)\,dθ\\ &(5)=\frac{-5^4}8\left(θ-\frac14\sin4θ\right)+C\\ &\qquad\left(\sin4θ=2\sin2θ\cos2θ=4\sinθ\cosθ\left(\cos^2θ-\sin^2θ\right)\right)\\ &(6)=\frac{-5^4}8\left(θ-\sinθ\cosθ\left(\cos^2θ-\sin^2θ\right)\right)+C\\ &\qquad\left(\cosθ=\frac x5\implies\sinθ=\frac{\sqrt{25-x^2}}5\right)\\ &(7)=\frac{-5^4}8\left(\arccos\left(\frac x5\right)-\frac{\sqrt{25-x^2}}5*\frac x5*\frac{25-2x^2}{5^2}\right)+C\\ &(8)=\frac{-5^4}8\left(\arccos\left(\frac x5\right)-\frac{x\sqrt{25-x^2}}{25}*\frac{25-2x^2}{25}\right)+C\\ \end{align}

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closed as unclear what you're asking by kingW3, Tom-Tom, Saad, Mohammad Riazi-Kermani, Xander Henderson Mar 7 '18 at 1:01

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    $\begingroup$ You've been told to use MathJax before, and you've been here long enough to know that mathematics in questions is expected to be formatted that way. You know that the question was all but impossible to read the way you originally posted it. $\endgroup$ – Robert Howard Mar 6 '18 at 20:40
  • $\begingroup$ Yeah sorry, I have tried it and it was a mess, so I just gave up. Will try to do it right now by looking at edits. $\endgroup$ – Stallmp Mar 6 '18 at 20:46
  • $\begingroup$ It can be a steep learning curve at first, but it just takes practice. I appreciate that you're making the effort to learn, however; that's more than can be said for some other new users. $\endgroup$ – Robert Howard Mar 6 '18 at 20:50
  • $\begingroup$ In the first line you have replaced $x^2$ by $(5\cos \theta)$ rather than by $(5\cos \theta)^2$. So the rest is probably wrong as well. $\endgroup$ – hardmath Mar 6 '18 at 22:32
  • $\begingroup$ Oh, that's a typo but I used (5cos theta)^2 correctly in my calculation $\endgroup$ – Stallmp Mar 6 '18 at 22:43
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$-5^4\int \sin^2\theta\cos^2\theta \ d\theta \\ -\frac {5^4}{4}\int \sin^2 2\theta \ d\theta\\ -\frac {5^4}{8}\int 1 - \cos 4\theta \ d\theta\\ -\frac {5^4}{8}(\theta - \frac 14 \sin 4\theta) + C\\ -\frac {5^4}{8}(\arccos \frac {\theta}{5} - \frac 14 (2\sin 2\theta\cos 2\theta)) + C\\ -\frac {5^4}{8}(\arccos \frac {\theta}{5} - \sin \theta\cos\theta(2\cos^2\theta - 1)) + C\\ -\frac {5^4}{8}(\arccos \frac {\theta}{5} - \sqrt {1-\frac {x^2}{25}}(\frac {x}{5})(2\frac {x^2}{25} - 1)) + C\\ \frac {1}{8} (-5^4\arccos \frac {x}{5} + (\sqrt {25- x^2})(x)(2x^2 - 25)) + C\\ $

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  • $\begingroup$ cos2θ=cos^2θ-sin^2θ right? How do you get 2cos^2θ-1? And aren't you supposed to replace arccos theta/5 with arccos x/5 ? $\endgroup$ – Stallmp Mar 6 '18 at 21:18
  • $\begingroup$ Also when I enter that antiderivative, it doesn't seem to be correct. $\endgroup$ – Stallmp Mar 6 '18 at 21:30
  • $\begingroup$ $\cos^2\theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1-2\sin^2\theta$ $\endgroup$ – Doug M Mar 6 '18 at 22:06
  • $\begingroup$ But how come if I enter F(2) in the formula, I get -110 (approximately)? When I use sin-substitution, I get a normal answer which is correct, but I always get a weird one using cos-substitution. The correct answer of F (2) should be 12,6 which is the same answer I get with the sin-sub. $\endgroup$ – Stallmp Mar 6 '18 at 22:14
  • $\begingroup$ $\frac {1}{8} (-625\arccos \frac {x}{5} + (\sqrt {25- x^2})(x)(2x^2 - 25)) + C = \frac {1}{8} (625\arcsin \frac {x}{5} + (\sqrt {25- x^2})(x)(2x^2 - 25)) + C$ $\endgroup$ – Doug M Mar 6 '18 at 22:31

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