Mathematics

# Integrate $\sqrt { 1-2x } dx$

##### SOLUTION
$\displaystyle I=\int \sqrt{1-2x}dx$
Substitute $\sqrt{1-2x}=t\Rightarrow 1-2x=t^2\Rightarrow -2dx=2tdt\Rightarrow -dx=tdt$
$\displaystyle I=\int t(-tdt)=-\int t^2dt=-\dfrac{t^3}{3}+C$
Undo $t=\sqrt{1-2x}$
$\displaystyle I=-\dfrac{(\sqrt{1-2x})^3}{3}+C$
$\displaystyle I=-\dfrac{(1-2x)^{3/2}}{3}+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Single Correct Medium
Solve :
$\displaystyle \int \dfrac {dx}{\sqrt {x - x^{2}}}$
• A. $2\sin^{-1} x + c$
• B. $2x \sin^{-1} x + c$
• C. $\sin^{-1} \sqrt {x} + c$
• D. $2\sin^{-1} \sqrt {x} + c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Fill in the blanks:
The value of $\displaystyle \int_{-\pi}^{\pi} sin^3 \,x cos^2 \,xdx$ is .........

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate $\displaystyle \int_{0}^{3}x^2 \ dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate : $\displaystyle \int \dfrac{dx}{\sin x+\sin 2x}$.

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$