First slide

Evaluation of definite integrals

Question

$\int_{\frac{1}{\sqrt{3}}}^{1} \frac{{(\tan^{- 1} x)}^{3}}{1 + x^{2}} d x =$

Easy

A

$\frac{1}{4} {(\frac{π}{12})}^{6}$
B

$\frac{1}{4} {(\frac{π}{12})}^{4}$
C

$\frac{1}{4} ({(\frac{π}{4})}^{4} - {(\frac{π}{6})}^{4})$
D

${(\frac{π}{8})}^{4}$

Solution

$\int_{π / 6}^{π / 4} t^{3} . d t$ $T a n^{- 1} x = t$
$= {(\frac{t^{4}}{4})}_{π / 6}^{π / 4}$

$= \frac{1}{4} (\frac{π^{4}}{4^{4}} - \frac{π^{4}}{6^{4}})$

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