Question
Evaluate the following integrals:
$\int_{0}^\limits{\text{a}}\sqrt{\text{a}^2-\text{x}^2}\text{ dx}$
$\int_{0}^\limits{\text{a}}\sqrt{\text{a}^2-\text{x}^2}\text{ dx}$
✓
Answer
Let $\text{x}=\text{a}\sin\theta$
Differentiating w.r.t. x, we get
$\text{dx}=\text{a}\cos\theta\text{ d}\theta$
Now, $\text{x}=0\Rightarrow\theta=0$
$\text{x}=\text{a}\Rightarrow\theta=\frac{\pi}{2}$
$\therefore\ \int_{0}^\limits{\text{a}}\sqrt{\text{a}^2-\text{x}^2}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\text{a}^2(1-\sin^2\theta)}\text{a}\cos\theta\text{ d}\theta$
$=\text{a}^2\int_{0}^\limits{\frac{\pi}{2}}\cos^2\theta\text{ d}\theta$ $\Big[\because(1-\sin^2\theta)=\cos^2\theta\text{ and }\frac{1+\cos2\theta}{2}=\cos2\theta\Big]$
$=\frac{\text{a}^2}{2}\int_{0}^\limits{\frac{\pi}{2}}\big(1+\cos2\theta\big)\text{d}\theta$
$=\frac{\text{a}^2}{2}\Big[\frac{\pi}{2}+0-0-0\big]$
$=\frac{\pi\text{a}^2}{4}$
Differentiating w.r.t. x, we get
$\text{dx}=\text{a}\cos\theta\text{ d}\theta$
Now, $\text{x}=0\Rightarrow\theta=0$
$\text{x}=\text{a}\Rightarrow\theta=\frac{\pi}{2}$
$\therefore\ \int_{0}^\limits{\text{a}}\sqrt{\text{a}^2-\text{x}^2}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\text{a}^2(1-\sin^2\theta)}\text{a}\cos\theta\text{ d}\theta$
$=\text{a}^2\int_{0}^\limits{\frac{\pi}{2}}\cos^2\theta\text{ d}\theta$ $\Big[\because(1-\sin^2\theta)=\cos^2\theta\text{ and }\frac{1+\cos2\theta}{2}=\cos2\theta\Big]$
$=\frac{\text{a}^2}{2}\int_{0}^\limits{\frac{\pi}{2}}\big(1+\cos2\theta\big)\text{d}\theta$
$=\frac{\text{a}^2}{2}\Big[\frac{\pi}{2}+0-0-0\big]$
$=\frac{\pi\text{a}^2}{4}$
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
Find the equation of the plane which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 and whose x-intercept is twice its z-intercept.
Hence write the vector equation of a plane passing through the point (2, 3, –1) and parallel to the plane obtained above.
Hence write the vector equation of a plane passing through the point (2, 3, –1) and parallel to the plane obtained above.
Evaluate the following integrals:
$\int\limits^2_{-2}\frac{3\text{x}^3+2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
→$\int\limits^2_{-2}\frac{3\text{x}^3+2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
Using matrices, solve the following system of equations:
x + y + z = 6
x + 2z = 7
3x + y + z = 12
A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of
6,000. Three times the award money for Hard work added to that given for honesty amounts to
11,000. The award money given for Honesty and Hardwork together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.
6,000. Three times the award money for Hard work added to that given for honesty amounts to11,000. The award money given for Honesty and Hardwork together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.Evaluate the following integrals:
$\int\frac{1}{\sin^4\text{x}+\sin^2\text{x}\cos^2\text{x}+\cos^4\text{x}}\ \text{dx}$
→$\int\frac{1}{\sin^4\text{x}+\sin^2\text{x}\cos^2\text{x}+\cos^4\text{x}}\ \text{dx}$
Find the coordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, -1, 3) and C(2, -3, -1).
Find the shortest distance between the following pairs of lines whose cartesian equation are:
$\frac{\text{x}-1}{-1}=\frac{\text{y}+2}{1}=\frac{\text{z}-3}{-2}$ and $\frac{\text{x}-1}{1}=\frac{\text{y}+1}{2}=\frac{\text{z}+1}{-2}$
→$\frac{\text{x}-1}{-1}=\frac{\text{y}+2}{1}=\frac{\text{z}-3}{-2}$ and $\frac{\text{x}-1}{1}=\frac{\text{y}+1}{2}=\frac{\text{z}+1}{-2}$
Using vector method, prove that the point is collinear:
A(6, -7, -1), B(2, -3, 1) and C(4, -5, 0)
A(6, -7, -1), B(2, -3, 1) and C(4, -5, 0)
Evaluate the following definite integrals:
$\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}$
→$\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}$
Let f: N → N be defined by ${f(n)}=\begin{cases}\frac{\text{n}+1}{2},\text{ if n is odd}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for all}\text{ n }\in\text{ N}.\\\frac{\text{n}}{2}\ \ \ \ ,\text{if n is even}\end{cases}$ State whether the function f is bijective. Justify your answer.
→