First slide

Evaluation of definite integrals

Question

_{$\int_{0}^{1} \frac{x^{3}}{{(1 + x^{2})}^{3}} d x =$}

Easy

A

_{$\frac{1}{8}$}
B

_{$\frac{1}{4}$}
C

_{$\frac{1}{12}$}
D

_{$\frac{1}{16}$}

Solution

$\int_{0}^{π / 4} \frac{T a n^{3} θ}{{(S e c^{2} θ)}^{3}} \times S e c^{2} θ d θ$       $\begin{matrix} x = \tan θ, \\ d x = \sec^{2} d θ \end{matrix}$
    $= \int_{0}^{π / 4} \frac{T a n^{3} θ}{S e c^{4} θ} d θ$
   $= \int_{0}^{π / 4} S i n^{3} θ . C o s θ d θ$
   $= {(\frac{S i n^{4} θ}{4})}_{0}^{π / 4}$
    $\frac{{(\frac{1}{\sqrt{2}})}^{4}}{4} = \frac{1}{16}$

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