First slide

Evaluation of definite integrals

Question

$\int_{1}^{16} \tan^{- 1} \sqrt{\sqrt{x} - 1} d x =$

Moderate

A

$\frac{8 π}{3} + \sqrt{3}$
B

$\frac{8 π}{3} - \sqrt{3}$
C

$\frac{16 π}{3} - 2 \sqrt{3}$
D

$\frac{16 π}{3} + 2 \sqrt{3}$

Solution

$\sqrt{x} - 1 = t^{2}, x = {(t^{2} + 1)}^{2}$
$I = \int_{0}^{\sqrt{3}} \tan^{- 1} t d {(t^{2} + 1)}^{2}$ $= \tan^{- 1} t {\cdot {(t^{2} + 1)}^{2} |}_{0}^{\sqrt{3}} - \int_{0}^{\sqrt{3}} (t^{2} + 1) d t$ by parts
$= \frac{16 π}{3} - {(\frac{t^{3}}{3} + t)}_{0}^{\sqrt{3}} = \frac{16 π}{3} - 2 \sqrt{3}$

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