Evaluate $$\int_{-7}^{-5}\frac{2}{y^4\sqrt{y^2-25}}dy$$

My attempt,

Let $y=5\sec \theta$

$dy=5\sec \theta \tan \theta d\theta$

$$=2\int_{\sec^{-1}(\frac{-7}{5})}^{\sec^{-1}(-1)}\frac{5\sec \theta \tan \theta}{625\sec^4 \theta \sqrt{25\sec^2 \theta -25}}d\theta$$

$$=2\int_{\sec^{-1}(\frac{-7}{5})}^{\sec^{-1}(-1)} \frac{\tan \theta}{\sec^3 \theta \cdot 5 \tan \theta} d\theta$$

$$=\frac{2}{625}2\int_{\sec^{-1}(\frac{-7}{5})}^{\sec^{-1}(-1)} \cos^3 \theta d\theta$$

Let $u=\sin \theta$

$du=\cos \theta d \theta$

$$=\frac{2}{625} \int_{\frac{2\sqrt{6}}{7}}^{0}1-u^2 du$$


But the given answer is in positive. Where have I done wrong? Hope someone can clarify for it. Thanks in advance.


I think the problem is when you evaluate the squareroot $\sqrt{25\tan^2\theta}$ and you simplify it to $5\tan \theta$ instead of $|5\tan \theta|$. You were assuming that $\tan \theta$ is always positive on the interval of integration. In fact, $\tan \theta$ is always negative on your interval, which explains the incorrect sign.

  • $\begingroup$ How to know tan \theta is negative on my interval? @riley $\endgroup$ – Mathxx Oct 18 '17 at 3:44
  • $\begingroup$ Well, we see that by the bounds of your integral, $\theta$ is always in the second quadrant. And $\tan\theta $ is always negative for angles in the second quadrant. $\endgroup$ – Riley Oct 18 '17 at 3:57

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