Mathematics

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

##### SOLUTION
Put x = sint   $\Rightarrow \frac{{dx}}{{dt}} = \cos tdt$

$\therefore I = \int_{}^{} {\frac{{ - \cos tdt}}{{\sqrt {1 - {{\sin }^2}t} }} = \int_{}^{} {\frac{{ - \cos tdt}}{{{{\cos }^2}t}}} }$

$= - t = {\sin ^{ - 1}}x$

${\cos ^{ - 1}}x + {\sin ^{ - 1}}x = \frac{\pi }{2}$

$\Rightarrow {\cos ^{ - 1}} - \frac{\pi }{2} = - {\sin ^{ - 1}}x$

$\therefore t = {\cos ^{ - 1}}x - \frac{\pi }{2}$

Here, $C = - \frac{\pi }{2}$

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Subjective Easy Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

Q1 Subjective Hard
Evaluate $\displaystyle \int_1^2 \dfrac{\sqrt x}{\sqrt{3 - x} + \sqrt x} dx$.

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The value of $\displaystyle \int\dfrac{3+2\cos x}{(2+3\cos x)^{2}}dx$ is equal to
• A. $\displaystyle \dfrac{2\cos x}{3\sin x+2}+c$
• B. $\dfrac{2\cos x}{3\cos x+2}+c$
• C. $\displaystyle \dfrac{2\sin x}{3\sin x+2}+c$
• D. $\displaystyle \dfrac{\sin x}{3\cos x+2}+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int \frac {\cos 2x}{(\sin x+\cos x)^2}dx$ is equal to
• A. $\dfrac {-1}{(\sin x+\cos x)}+C$
• B. $\log |\sin x-\cos x|+C$
• C. $\dfrac {1}{(\sin x+\cos x)^2}$
• D. $\log |\sin x+\cos x|+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate the following integral:
$\displaystyle\int^{\pi/2}_0\dfrac{\sin x\cos x}{(\cos^2x+3\cos x+2)}dx$.

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
Evaluate: $\displaystyle \int { \dfrac { \cos { x } -\sin { x } }{ 1+\sin { 2x } } } dx$
• A. $\dfrac{1}{\sin x+\cos x}+C$
• B. $\dfrac{2}{\sin 2x+\cos x}+C$
• C. $-\dfrac{2}{\sin 2x+\cos x}+C$
• D. $-\dfrac{1}{\sin x+\cos x}+C$