Mathematics

# Solve : $\underset{0}{\overset{1}{\int}} \sqrt{9 - 4x^2} dx$

##### SOLUTION
$\int \sqrt{9-4x^2}dx$

$x=\dfrac{3}{2}\sin u\rightarrow dx=\dfrac{3}{2}\cos u du$

$=\int \left ( \sqrt{9-4\frac{9}{4}\sin^2 u} \right )\frac{3}{2}\cos u du$

$=\int \sqrt{9(1-\sin^2 u)}\frac{3}{2}\cos u du$

$=\int (3\cos u) \frac{3}{2}\cos u du$

$=\dfrac{9}{2}\int \cos ^2 u du$

$=\dfrac{9}{2}\int\left ( \frac{1+\cos 2u}{2} \right )du$

$=\dfrac{9}{2}\cdot \dfrac{1}{2}\left ( u+\dfrac{1}{2}\sin u \right )$

$u=\sin^{-1}\left ( \dfrac{2}{3}x \right )$

$\int \sqrt{9-4x^2}dx=\dfrac{9}{2}\cdot \dfrac{1}{2}\left ( \sin^{-1}\left ( \dfrac{2}{3}x \right )+\dfrac{1}{2}\sin\left ( \sin^{-1}\left ( \dfrac{2}{3}x \right ) \right )\right )+C$

$\int_{0}^{1}\sqrt{9-4x^2}dx=\dfrac{9}{4}\sin^{-1}\left ( \dfrac{2}{3} \right )+\dfrac{\sqrt{5}}{2}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Hard
The value of the integral $\displaystyle \int \frac{(1-\cos\theta )^{2/7}}{(1+\cos\theta )^{9/7}}d\theta$ is
• A. $\dfrac{7}{11}\left (\cos\dfrac{\theta }{2}\right)^{\dfrac{11}{7}}+C$
• B. $\dfrac{7}{11}\left (\sin\dfrac{\theta }{2}\right)^{\dfrac{11}{7}}+C$
• C. None of these
• D. $\displaystyle \frac{7}{11}\left (\tan\frac{\theta }{2}\right)^{\dfrac{11}{7}}+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate the following integrals:
$\int { \sin ^{ 3 }{ x } \cos ^{ 5 }{ x } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
$\int\frac{tanx+tan^3x}{1+tanx}dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Find $F (x)$ from the given $F'(x)$
$F'(x) = 4x + 1$ and $F(-1) = 2$.

The value of $\left(\int_{0}^{\pi/6} sec^{2} x dx\right)^2$ is: