Mathematics

# Evaluate the following integrals:$\int { \sqrt { x-{ x }^{ 2 } } } dx$

##### SOLUTION
Let $I=\displaystyle\int \sqrt{x-x^2}\ d x$

$\implies I=\displaystyle\int \sqrt{-\left(-2\bigg(\dfrac{1}{2}\bigg)(x)+x^2\right)}\ d x$

$\implies I=\displaystyle\int \sqrt{\bigg(\dfrac{1}{2}\bigg)^2-\left(\bigg(\dfrac{1}{2}\bigg)^2-2(a)(x)+x^2\right)}\ d x$

$\implies I=\displaystyle\int \sqrt{\dfrac{1}{4}-\left(\dfrac{1}{2}-x\right)^2}\ d x$

Put $t=a-x\implies d t=-d x$

$\implies I=-\displaystyle\int \sqrt{\bigg(\dfrac{1}{2}\bigg)^2-t^2}\ d t$

As we know that

$\displaystyle\int \sqrt{a^2-x^2}\ d x=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\text{sin}^{-1} \left(\dfrac{x}{a}\right)+C$

$I=-\dfrac{t}{2}\sqrt{\dfrac{1}{4}-t^2}-\dfrac{1/4}{2}\text{sin}^{-1} \left(\dfrac{t}{1/2}\right)+C$

$I=-\dfrac{\dfrac{1}{2}-x}{2}\sqrt{\dfrac{1}{4}-\bigg(\dfrac{1}{2}-x\bigg)^2}-\dfrac{1/4}{2}\text{sin}^{-1} \left(2\bigg(x-\dfrac{1}{2}\bigg)\right)+C$

$I=\dfrac{1}{4}(2 x-1)\sqrt{x-x^2}-\dfrac{1}{8}\text{sin}^{-1} \left(2 x-1\right)+C$

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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 Medium
The value of the definite integral $\int _{ -1 }^{ 1 }{ x } In({ 1 }^{ x }+{ 2 }^{ x }+{ 3 }^{ x }+{ 6 }^{ x })dx$ equals
• A. $\frac { In2+In3 }{ 2 }$
• B. $In2+In3$
• C. $\frac { In2+In3 }{ 4 }$
• D. $\frac { In2+In3 }{ 3 }$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Solve $\displaystyle \int_0^1\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}dx$ is equal to
• A. $\pi/2$
• B. $\pi/4$
• C. $0$
• D. $1$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int_{0}^{\dfrac{\pi}{2}}$ $\displaystyle \frac{200\sin x+100\cos x}{\sin x+\cos x}dx=$
• A. $50$ $\pi$
• B. $25$ $\pi$
• C. $150$ $\pi$
• D. $75$ $\pi$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate : $\int { \sqrt { \dfrac { x }{ { x }^{ 3 }{ a }^{ 3 } } } dx }$

$\int \dfrac{\sin x}{\sqrt 3+\cos x}dx$