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Nesterboy [21]
3 years ago
5

Evaluate the integral using the indicated trigonometric substitution. (use c for the constant of integration.) x^3 / sqrt x^2 +

49 dx , x = 7 tan(θ)
Mathematics
1 answer:
slava [35]3 years ago
6 0
\displaystyle\int\frac{x^3}{\sqrt{x^2+49}}\,\mathrm dx

Taking x=7\tan\theta gives \mathrm dx=7\sec^2\theta\,\mathrm d\theta, so that the integral becomes

\displaystyle\int\frac{(7\tan\theta)^3}{\sqrt{(7\tan\theta)^2+49}}(7\sec^2\theta)\,\mathrm d\theta
=\displaystyle7^4\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{49\tan^2\theta+49}}\,\mathrm d\theta
=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\tan^2\theta+1}}\,\mathrm d\theta
=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\sec^2\theta}}\,\mathrm d\theta
=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{|\sec\theta|}\,\mathrm d\theta

When \sec\theta>0, we have

=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sec\theta}\,\mathrm d\theta
=\displaystyle7^3\int\tan^3\theta\sec^2\theta\,\mathrm d\theta

and from here we can substitute u=\tan\theta to proceed from here.

Quick note: When we set x=7\tan\theta, we are implicitly enforcing -\dfrac\pi2 just so that the substitution can be undone later via \theta=\tan^{-1}\dfrac x7. But note that over this domain, we automatically guarantee that \sec\theta>0, so the absolute value bars can be dropped immediately.
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