Mathematics

# Integrate $\int {\sqrt {1 - {t^2}} } dt$.

##### SOLUTION
$\displaystyle \int \sqrt{1-t^{2}} dt$

let $t = sin\theta \Rightarrow \theta = sin^{-1}(t)$ and $cos\theta =\sqrt{1-sin^{2}\theta}$

$= \sqrt{1-t^{2}}$
$dt = cos\theta d\theta$

$\displaystyle\int \sqrt{1-sin^{2}\theta} \,\,cos\theta d\theta = \int cos\theta . cos\theta .d\theta$
$\displaystyle= \int cos^{2}\theta d\theta$

$\displaystyle = \int (\dfrac{1+cos2\theta}{2}) d\theta$
$\displaystyle= \dfrac{1}{2}\int d\theta+\dfrac{1}{2}\int cos2\theta d\theta$

$\displaystyle = \dfrac{1}{2}\theta + \dfrac{1}{2}.\dfrac{sin2\theta}{2}+C$
$\displaystyle = \dfrac{\theta}{2}+\dfrac{sin2\theta}{4}+c$

$\displaystyle = \dfrac{1}{2}sin^{-1}(t)+\dfrac{2sincos\theta}{4}+c$

$\displaystyle = \dfrac{1}{2}sin^{-1}(t)+\dfrac{1}{2}.t\sqrt{1-t^{2}}+c$

$\displaystyle \int \sqrt{1-t^{2}}dt = \dfrac{1}{2} sin^{-1}(t)+\frac{t}{2}\sqrt{1-t^{2}}+c$         (where $t=\sin\theta$)

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Subjective Hard
Prove that $\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ 2 } { \tan }^{ 3 }x dx =1-\log{2}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard

Integrate
by using a suitable substitution:-

(a) $\int {{{\rm{3}} \over {{{\left( {{\rm{2 - x}}} \right)}^{\rm{2}}}}}{\rm{dx}}}$

(b) ${\rm{\;\;}}\int {{\rm{sin}}\left( {{\rm{8z - 5}}} \right){\rm{dz}}}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $f(x)={ \{ }_{ 0, \quad \quad \quad |x|>1\quad \quad \quad \quad \quad \quad \quad \quad \quad }^{ 1-|x|,\quad \quad |x|<1 }$ and $g(x)=f(x-1)+f(x+1)$ then $\int _{ 0 }^{ 3 }{ g(x)dx }$ is equal to
• A. 1
• B. 2
• C. 3
• D.

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int \cot x {dx}=$
• A. $\ln (\sin^2x) +C$
• B. $(\sin x) +C$
• C. None of these
• D. $\ln (\sin x) +C$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
$\displaystyle \int\sqrt{\frac{\cos x-\cos^{3}x}{1-\cos^{3}x}}dx=$
• A. $\displaystyle \frac{2}{3}\sin^{-1}(\cos^{\frac{3}{2}}x)+c$
• B. $\displaystyle \frac{3}{2}\sin^{-1}(\cos^{\frac{3}{2}}x)+c$
• C. $\displaystyle \frac{3}{2} \cos^{-1} (\cos^{\frac{3}{2}}x)+c$
• D. $\displaystyle \frac{2}{3} \cos^{-1} (\cos^{\frac{3}{2}}x)+c$