Show that - x3-q2x8-p2dx-1-4q - t-2-t2-p2-1-2p2log t-t2-p2 - where t-qx4-

Let I=∫ #160;x^ 3 √( q^ 2 x^ 8 p^ 2 ) dx Put qx^ 4 =t ⇒ 4qx^ 3 dx=dt I= 1 / 4q ∫ #160;√( t^ 2 p^ 2 ) dt = 1 4q [ t 2 √( t^ 2 ...

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Byju's Answer

Standard XII

Mathematics

Special Integrals - 1

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Question

Show that $\int x^{3} \sqrt{q^{2} x^{8} - p^{2}} d x = \frac{1}{4 q} [\frac{t}{2} \sqrt{t^{2} - p^{2}} - \frac{1}{2} p^{2} log (t + \sqrt{t^{2} - p^{2}})] .$ where $t = q x^{4}$ .

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Solution

Let $I = \int x^{3} \sqrt{q^{2} x^{8} - p^{2}} d x$

Put $q x^{4} = t$
$\Rightarrow 4 q x^{3} d x = d t$

$I = \frac{1}{4 q} \int \sqrt{t^{2} - p^{2}} d t$
$= \frac{1}{4 q} [\frac{t}{2} \sqrt{t^{2} - p^{2}} - \frac{1}{2} p^{2} log (t + \sqrt{t^{2} - p^{2}})]$ where $t = q x^{4}$ .

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