Question
Evaluate : $\int \sqrt{a^2-x^2} \cdot d x$
✓
Answer
Let $\mathrm{I}=\int \sqrt{a^2-x^2} \cdot 1 \cdot d x$
$
\begin{aligned}
& =\sqrt{a^2-x^2} \cdot \int 1 \cdot d x-\int \frac{d}{d x} \cdot \sqrt{a^2-x^2} \cdot \int 1 \cdot d x \cdot d x \\
& =\sqrt{a^2-x^2} \cdot x-\int \frac{1}{2 \sqrt{a^2-x^2}}(-2 x) \cdot(x) \cdot d x \\
& =\sqrt{a^2-x^2} \cdot x+\int \frac{x^2}{\sqrt{a^2-x^2}} \cdot d x \\
& =\sqrt{a^2-x^2} \cdot x+\int \frac{a^2-\left(a^2-x^2\right)}{\sqrt{a^2-x^2}} \cdot d x \\
& =\sqrt{a^2-x^2} \cdot x+\int\left[\frac{a^2}{\sqrt{a^2-x^2}}-\frac{\left(a^2-x^2\right)}{\sqrt{a^2-x^2}}\right] \cdot d x \\
& =x \cdot \sqrt{a^2-x^2}+a^2 \int \frac{1}{\sqrt{a^2-x^2}} \cdot d x-\int \sqrt{a^2-x^2} \cdot d x \\
& \mathrm{I}=x \cdot \sqrt{a^2-x^2}+a^2 \int \frac{1}{\sqrt{a^2-x^2}} \cdot d x-\mathrm{I} \\
& \therefore \quad \mathrm{I}+\mathrm{I} \quad=x \cdot \sqrt{a^2-x^2}+a^2 \cdot \sin ^{-1}\left(\frac{x}{a}\right)+c \\
& \therefore \quad \text { I } \quad=\frac{x}{2} \cdot \sqrt{a^2-x^2}+\frac{a^2}{2} \cdot \sin ^{-1}\left(\frac{x}{a}\right)+c \\
& \therefore \quad \int \sqrt{a^2-x^2} \cdot d x=\frac{x}{2} \cdot \sqrt{a^2-x^2}+\frac{a^2}{2} \cdot \sin ^{-1}\left(\frac{x}{a}\right)+c \\
& \text { e.g. } \int \sqrt{9-x^2} \cdot d x=\frac{x}{2} \cdot \sqrt{9-x^2}+\frac{9}{2} \cdot \sin ^{-1}\left(\frac{x}{3}\right)+c \\
&
\end{aligned}
$
with reference to the above example solve these :
$
\begin{aligned}
& =\sqrt{a^2-x^2} \cdot \int 1 \cdot d x-\int \frac{d}{d x} \cdot \sqrt{a^2-x^2} \cdot \int 1 \cdot d x \cdot d x \\
& =\sqrt{a^2-x^2} \cdot x-\int \frac{1}{2 \sqrt{a^2-x^2}}(-2 x) \cdot(x) \cdot d x \\
& =\sqrt{a^2-x^2} \cdot x+\int \frac{x^2}{\sqrt{a^2-x^2}} \cdot d x \\
& =\sqrt{a^2-x^2} \cdot x+\int \frac{a^2-\left(a^2-x^2\right)}{\sqrt{a^2-x^2}} \cdot d x \\
& =\sqrt{a^2-x^2} \cdot x+\int\left[\frac{a^2}{\sqrt{a^2-x^2}}-\frac{\left(a^2-x^2\right)}{\sqrt{a^2-x^2}}\right] \cdot d x \\
& =x \cdot \sqrt{a^2-x^2}+a^2 \int \frac{1}{\sqrt{a^2-x^2}} \cdot d x-\int \sqrt{a^2-x^2} \cdot d x \\
& \mathrm{I}=x \cdot \sqrt{a^2-x^2}+a^2 \int \frac{1}{\sqrt{a^2-x^2}} \cdot d x-\mathrm{I} \\
& \therefore \quad \mathrm{I}+\mathrm{I} \quad=x \cdot \sqrt{a^2-x^2}+a^2 \cdot \sin ^{-1}\left(\frac{x}{a}\right)+c \\
& \therefore \quad \text { I } \quad=\frac{x}{2} \cdot \sqrt{a^2-x^2}+\frac{a^2}{2} \cdot \sin ^{-1}\left(\frac{x}{a}\right)+c \\
& \therefore \quad \int \sqrt{a^2-x^2} \cdot d x=\frac{x}{2} \cdot \sqrt{a^2-x^2}+\frac{a^2}{2} \cdot \sin ^{-1}\left(\frac{x}{a}\right)+c \\
& \text { e.g. } \int \sqrt{9-x^2} \cdot d x=\frac{x}{2} \cdot \sqrt{9-x^2}+\frac{9}{2} \cdot \sin ^{-1}\left(\frac{x}{3}\right)+c \\
&
\end{aligned}
$
with reference to the above example solve these :
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Verify Lagrange’s mean value theorem for the following functions: f(x) = (x – 1)(x – 2)(x – 3) on [0, 4]
Write the symbolic form of the following switching circuits construct its switching table and interpret it.
Evalution $\int \frac{5 e^x}{\left(e^x+1\right)\left(e^{2 x}+9\right)} d x$
→Find the inverse of the following matrices by the adjoint method : $\left[\begin{array}{rr}-1 & 5 \\ -3 & 2\end{array}\right]$
→Diffrentiate the following w. r. t. x.$\cot ^{-1}\left(\frac{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}+\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}-\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}\right)$
→If $A=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$ then find $A^{-1}$ by Adjoint method.
→If $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right]$, find $A B$ and $(A B)^{-1}$
→Find the derivative of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ w.r.t. $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$.
→Given the probability density function (p.d.f.) of a continuous random variable $x$ as:
$\begin{aligned}
f(x) & =\frac{x^2}{3}, & -1<x<2 \\
& =0, & \text { otherwise }
\end{aligned}$
Determine the cumulative distribution function (c.d.f.) of $X$ and hence find
$P ( X <1), P ( X >0), P (1< X <2) \text {. }$
→$\begin{aligned}
f(x) & =\frac{x^2}{3}, & -1<x<2 \\
& =0, & \text { otherwise }
\end{aligned}$
Determine the cumulative distribution function (c.d.f.) of $X$ and hence find
$P ( X <1), P ( X >0), P (1< X <2) \text {. }$
For the differential equation, find the particular solution
$\frac{ d y}{ d x}=(4 x + y +1), \text { when } y =1, x =0$
→$\frac{ d y}{ d x}=(4 x + y +1), \text { when } y =1, x =0$