Integrate the following functions w.r.t. x: ( ^ -1 x )^ 3 2 1-x^2 - Vidyadip

Question

Integrate the following functions w.r.t. x:

$\frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^2}}$

✓

Answer

Let $I =\int \frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^2}} d x$

Put $\sin ^{-1} x=t . \quad \therefore \frac{1}{\sqrt{1-x^2}} d x=d t$

$\therefore I=\int t^{\frac{3}{2}} d t=\frac{t^{\frac{5}{2}}}{5 / 2}+c$

$=\frac{2}{5}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c$

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

Explore more

More "Answer the following questions in short." from Indefinite Integration→Indefinite Integration — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

Show that the function $f(x)=x^3+10 x+7$ for $x \in \mathrm{R}$ is strictly increasing.

Find the separate equations of the lines represented by following equations: $3 x^2-10 x y-8 y^2=0$

Find mean for the following probability distribution.

X	0	1	2	3
P(X=x)	$\frac{1}{6}$	$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{6}$

Check the validity of the Rolle’s theorem for the following functions.

$f(x)=x^{\frac{2}{3}}, x \in[-1,1]$

Using rules of negation write the negations of the following with justification. ~q → p

Which of the following equations have no solutions :

$\cos ^2 \theta=-1$

Write the negations of the following stating the rules used : $(p \rightarrow q) \vee r$

$\int \sqrt{1-\cos 2 x} \cdot d x$

Write the degree of the differential equation $\frac{\text{d}^{2}\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}=2\text{x}^{2}\log\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^2}\Big).$

Let $X$ be a continuous random variable whose probability density function is $f(x)=3 x^2$, for $0<x<1$. note that $f(x)$ is not $P[X=x]$.
For example, $f(0.9)=3(0.9)^2=2 \cdot 43>1$, which is clearly not a probability. In the continuous case, $f(x)$ is the height of the curve at $X=x$, so that the total area under the curve is 1 . Here it is areas under the curve that define the probabilities.