Question
Integrate the function $\frac{x^{3}}{\sqrt{1-x^{8}}}$

Answer

Let $I=\frac{x^{3}}{\sqrt{1-x^{8}}}$ 
Now, let x4 = t $\Rightarrow$ 4x3 dx = dt
And x3 dx = $\frac{dt}{4}$ 
$\Rightarrow \int \frac{\mathrm{x}^{3}}{\sqrt{1-\mathrm{x}^{8}}} \mathrm{dx}=\int \frac{1}{\sqrt{1-\mathrm{t}^{2}}}\left(\frac{\mathrm{dt}}{4}\right)$ 
$=\frac{1}{4} \int \frac{1}{\sqrt{1^{2}-t^{2}}} \cdot d t$ 
= $\frac{1}{4} \sin ^{-1} t+C$ 
$\Rightarrow \mathrm{I}=\frac{1}{4} \sin ^{-1}\left(\mathrm{x}^{4}\right)+\mathrm{C}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Let f: {2, 3, 4, 5) $\rightarrow$ {3, 4 ,5, 9} and g : {3, 4, 5, 9) $\rightarrow$ {7, 11, 15) be functions defined as f (2) = 3 , f (3) = 4 , f (4) = f (5) = 5 and, g (3) = g (4) = 7 and g (5) = g (9) = 11. Find gof.
Integrate the function w.r.t. x: $\frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}}$
Write the direction cosines of the line joining the points (1, 0, 0) and (0, 1, 1).
Let R be a relation on the set A of ordered pairs of positive integers defined by
(x, y) R (u, v) if and only if xv = yu.
Show that R is an equivalence relation.
Find the principal value of the following:
$\sin^{-1}\Big(\tan\frac{5\pi}{4}\Big)$
The total cost C(x) in Rupees, associated with the production of x units of an item is given by C(x) = 0.005x3 - 0.02x2 + 30x + 5000.
Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output.
Write the equation of a plane which is at a distance of $5\sqrt{3}$ units from origin and the normal to which is equally inclined to coordinate axes.
Find $\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x$ 
The total cost C(x) in rupees associated with the production of x units of an item given by c(x) = 0.007x3 - 0.003x2 + 15x + 4000. Find the marginal cost when 17 units are produced.
Find the Cartesian equation of the line which passes through the point (–2, 4, –5) and is parallel to the line $\frac{\text{x} + 3 }{3} =\frac{4 - \text{y}}{5} = \frac{\text{z} + 8}{6}.$