Mathematics

# Evaluate the following integrals:$\int { \sqrt { 1+x-2{ x }^{ 2 } } } dx$

##### SOLUTION
Let $I=\displaystyle\int \sqrt{1+x-2 x^2} d x$

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

$=\displaystyle\int \sqrt{1+\dfrac{1}{8}-2\left(\bigg(\dfrac{1}{4}\bigg)^2-2(x)\bigg(\dfrac{1}{4}\bigg)+x^2\right)}\ d x$

$=\sqrt{2}\displaystyle\int \sqrt{\dfrac{9}{16}-\bigg(x-\dfrac{1}{4}\bigg)^2}\ d x$                                                        $(\because (a-b)^2=a^2-2 a b+b^2)$

Put $t=x-\dfrac{1}{4}\implies d t=d x$

$I=\sqrt{2}\displaystyle\int \sqrt{\bigg(\dfrac{3}{4}\bigg)^2-t^2}\ d t$

As we know that

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

Here $a=\dfrac{3}{4}$

$\implies I=\sqrt{2}\left(\dfrac{t}{2}\sqrt{\dfrac{9}{16}-t^2}+\dfrac{9/16}{2}\text{sin}^{-1} \left(\dfrac{t}{3/4}\right)+C\right)$

$\implies I=\dfrac{1}{2}\bigg(x-\dfrac{1}{4}\bigg)\sqrt{\dfrac{9}{8}-2\bigg(x-\dfrac{1}{4}\bigg)^2}+\dfrac{9\sqrt{2}}{32}\text{sin}^{-1} \bigg(\dfrac{4(x-1/4)}{3}\bigg)+C$

$\implies I=\dfrac{1}{8}(4 x-1)\sqrt{1+x-2 x^2}+\dfrac{9\sqrt{2}}{32}\text{sin}^{-1} \bigg(\dfrac{4 x-1}{3}\bigg)+C$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Subjective Medium
Evaluate $\displaystyle\int^2_1\dfrac{dx}{x(1+2x)^2}$.

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Evaluate $\displaystyle \int\frac{\sin^{10}x}{\cos^{12}x}dx$
• A. $\displaystyle \frac{\tan^{10}x}{10}+c$
• B. $\displaystyle 10\tan^{9}x+c$
• C. $\displaystyle \frac{\tan^{9}x}{10}+c$
• D. $\displaystyle \frac{\tan^{11}x}{11}+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\alpha > 1$, then $\int \dfrac {dx}{x^{2} + 2\alpha x + 1} =$.
• A. $\dfrac {1}{\sqrt {1 - \alpha^{2}}}\cot^{-1} \left (\dfrac {x + \alpha}{\sqrt {1 - \alpha^{2}}}\right ) + c$
• B. $\dfrac {1}{2\sqrt {\alpha^{2} - 1}}\log \left (\dfrac {x + \alpha - \sqrt {\alpha^{2} - 1}}{x + \alpha + \sqrt {\alpha^{2} - 1}}\right ) + c$
• C. $\dfrac {1}{2\sqrt {\alpha^{2} - 1}}\log \left (\dfrac {x + \alpha + \sqrt {\alpha^{2} - 1}}{x + \alpha - \sqrt {\alpha^{2} - 1}}\right ) + c$
• D. $\dfrac {1}{\sqrt {1 - \alpha^{2}}}\tan^{-1} \left (\dfrac {x + \alpha}{\sqrt {1 - \alpha^{2}}}\right ) + c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
The value of intergral $\displaystyle \int_{\dfrac{\pi}{4}}^{\dfrac{3\pi}{4}}\dfrac{x}{1+\sin x}dx$ is :
• A. $\pi\sqrt{2}$
• B. $\dfrac{\pi}{2}(\sqrt{2}+1)$
• C. $2\pi (\sqrt{2}-1)$
• D. $\pi(\sqrt{2}-1)$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$